Class 11 Basic Mathematics – Derivatives and Trigonometric Functions
This chapter provides a comprehensive overview of Derivatives and Trigonometric Functions in Class 11 Basic Mathematics. Understanding these fundamental concepts is crucial for mastering calculus and advanced mathematical applications. These topics form the foundation for higher mathematics and various scientific disciplines.
Chapter Information
Chapter: Derivatives and Trigonometric Functions
Subject: Class 11 Basic Mathematics (NEB Syllabus)
Description: This guide provides complete Basic Mathematics notes covering derivatives of trigonometric functions, chain rule applications, and comprehensive solved problems as per NEB syllabus.
Credit: Important Notes
Table of Contents
Detailed Chapter Notes
Derivatives of Basic Trigonometric Functions
The derivatives of trigonometric functions are fundamental in calculus. Below are detailed derivations using first principles.
1. Derivative of $\sin(x)$
Solution:
Let $y = \sin(x)$
Let $\Delta x$ and $\Delta y$ be the small increments in $x$ and $y$ respectively.
Then, $y + \Delta y = \sin(x + \Delta x)$
or, $\Delta y = \sin(x + \Delta x) – \sin(x)$
or, $\Delta y = 2\cos\left(\frac{x + \Delta x + x}{2}\right) \cdot \sin\left(\frac{x + \Delta x – x}{2}\right)$
or, $\Delta y = 2\cos\left(x + \frac{\Delta x}{2}\right) \cdot \sin\left(\frac{\Delta x}{2}\right)$
or, $\frac{\Delta y}{\Delta x} = \frac{2\cos\left(x + \frac{\Delta x}{2}\right) \cdot \sin\left(\frac{\Delta x}{2}\right)}{\Delta x}$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \cos\left(x + \frac{\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}}$
or, $\frac{dy}{dx} = \cos(x) \cdot 1$
$\therefore \frac{d}{dx}(\sin(x)) = \cos(x)$
2. Derivative of $\cos(x)$
Solution:
Let $y = \cos(x)$
Let $\Delta x$ and $\Delta y$ be the small increments in $x$ and $y$ respectively.
Then, $y + \Delta y = \cos(x + \Delta x)$
or, $\Delta y = \cos(x + \Delta x) – \cos(x)$
or, $\Delta y = -2\sin\left(\frac{x + \Delta x + x}{2}\right) \cdot \sin\left(\frac{x + \Delta x – x}{2}\right)$
or, $\Delta y = -2\sin\left(x + \frac{\Delta x}{2}\right) \cdot \sin\left(\frac{\Delta x}{2}\right)$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} -\sin\left(x + \frac{\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}}$
or, $\frac{dy}{dx} = -\sin(x) \cdot 1$
$\therefore \frac{d}{dx}(\cos(x)) = -\sin(x)$
3. Derivative of $\tan(x)$
Solution:
Let $y = \tan(x)$
Let $\Delta x$ and $\Delta y$ be the small increments in $x$ and $y$ respectively.
Then, $y + \Delta y = \tan(x + \Delta x)$
or, $\Delta y = \tan(x + \Delta x) – \tan(x)$
or, $\Delta y = \frac{\sin(x + \Delta x)}{\cos(x + \Delta x)} – \frac{\sin(x)}{\cos(x)}$
or, $\Delta y = \frac{\sin(x + \Delta x)\cos(x) – \cos(x + \Delta x)\sin(x)}{\cos(x + \Delta x)\cos(x)}$
or, $\Delta y = \frac{\sin(x + \Delta x – x)}{\cos(x + \Delta x)\cos(x)} = \frac{\sin(\Delta x)}{\cos(x + \Delta x)\cos(x)}$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{\sin(\Delta x)}{\Delta x} \cdot \frac{1}{\cos(x + \Delta x)\cos(x)}$
or, $\frac{dy}{dx} = 1 \cdot \frac{1}{\cos(x)\cos(x)} = \frac{1}{\cos^2(x)}$
$\therefore \frac{d}{dx}(\tan(x)) = \sec^2(x)$
4. Derivative of $\cot(x)$
Solution:
Let $y = \cot(x)$
$\Delta y = \cot(x + \Delta x) – \cot(x) = \frac{\cos(x + \Delta x)}{\sin(x + \Delta x)} – \frac{\cos(x)}{\sin(x)}$
or, $\Delta y = \frac{\sin(x)\cos(x + \Delta x) – \cos(x)\sin(x + \Delta x)}{\sin(x + \Delta x)\sin(x)}$
or, $\Delta y = \frac{\sin(x – (x + \Delta x))}{\sin(x + \Delta x)\sin(x)} = \frac{\sin(-\Delta x)}{\sin(x + \Delta x)\sin(x)}$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{-\sin(\Delta x)}{\Delta x} \cdot \frac{1}{\sin(x + \Delta x)\sin(x)}$
or, $\frac{dy}{dx} = -1 \cdot \frac{1}{\sin^2(x)}$
$\therefore \frac{d}{dx}(\cot(x)) = -\csc^2(x)$
5. Derivative of $\csc(x)$
Solution:
Let $y = \csc(x)$
$\Delta y = \csc(x + \Delta x) – \csc(x) = \frac{1}{\sin(x + \Delta x)} – \frac{1}{\sin(x)}$
or, $\Delta y = \frac{\sin(x) – \sin(x + \Delta x)}{\sin(x + \Delta x)\sin(x)}$
or, $\Delta y = \frac{2\cos\left(\frac{x + x + \Delta x}{2}\right) \sin\left(\frac{x – x – \Delta x}{2}\right)}{\sin(x + \Delta x)\sin(x)}$
or, $\Delta y = \frac{2\cos\left(x + \frac{\Delta x}{2}\right) \sin\left(-\frac{\Delta x}{2}\right)}{\sin(x + \Delta x)\sin(x)}$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{-\cos\left(x + \frac{\Delta x}{2}\right)}{\sin(x + \Delta x)\sin(x)} \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}}$
or, $\frac{dy}{dx} = \frac{-\cos(x)}{\sin(x)\sin(x)} \cdot 1 = -\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)}$
$\therefore \frac{d}{dx}(\csc(x)) = -\csc(x)\cot(x)$
6. Derivative of $\sec(x)$
Solution:
Let $y = \sec(x)$
$\Delta y = \sec(x + \Delta x) – \sec(x) = \frac{1}{\cos(x + \Delta x)} – \frac{1}{\cos(x)}$
or, $\Delta y = \frac{\cos(x) – \cos(x + \Delta x)}{\cos(x + \Delta x)\cos(x)}$
or, $\Delta y = \frac{-2\sin\left(\frac{x + x + \Delta x}{2}\right) \sin\left(\frac{x – x – \Delta x}{2}\right)}{\cos(x + \Delta x)\cos(x)}$
or, $\Delta y = \frac{2\sin\left(x + \frac{\Delta x}{2}\right) \sin\left(\frac{\Delta x}{2}\right)}{\cos(x + \Delta x)\cos(x)}$
or, $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{\sin\left(x + \frac{\Delta x}{2}\right)}{\cos(x + \Delta x)\cos(x)} \cdot \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}}$
or, $\frac{dy}{dx} = \frac{\sin(x)}{\cos(x)\cos(x)} \cdot 1 = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}$
$\therefore \frac{d}{dx}(\sec(x)) = \tan(x)\sec(x)$
Exercise 16.2 – Applications and Problems
1. Find, from the first principles, the differential coefficient of:
i. $\sin(4x)$
Let $y = \sin(4x)$
$\Delta y = \sin(4(x + \Delta x)) – \sin(4x) = \sin(4x + 4\Delta x) – \sin(4x)$
or, $\Delta y = 2\cos\left(\frac{4x + 4\Delta x + 4x}{2}\right) \sin\left(\frac{4x + 4\Delta x – 4x}{2}\right)$
or, $\Delta y = 2\cos(4x + 2\Delta x) \sin(2\Delta x)$
or, $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{2\cos(4x + 2\Delta x) \sin(2\Delta x)}{\Delta x} = \lim_{\Delta x \to 0} 2\cos(4x + 2\Delta x) \cdot \frac{\sin(2\Delta x)}{2\Delta x} \cdot 2$
or, $\frac{dy}{dx} = 2\cos(4x) \cdot 1 \cdot 2$
$\therefore \frac{dy}{dx} = 4\cos(4x)$
ii. $\cos(ax – b)$
Let $y = \cos(ax – b)$
$\Delta y = \cos(a(x + \Delta x) – b) – \cos(ax – b)$
or, $\Delta y = -2\sin\left(\frac{ax + a\Delta x – b + ax – b}{2}\right) \sin\left(\frac{ax + a\Delta x – b – (ax – b)}{2}\right)$
or, $\Delta y = -2\sin\left(ax – b + \frac{a\Delta x}{2}\right) \sin\left(\frac{a\Delta x}{2}\right)$
or, $\frac{dy}{dx} = \lim_{\Delta x \to 0} -2\sin\left(ax – b + \frac{a\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{a\Delta x}{2}\right)}{\Delta x}$
or, $\frac{dy}{dx} = \lim_{\Delta x \to 0} -\sin\left(ax – b + \frac{a\Delta x}{2}\right) \cdot \frac{\sin\left(\frac{a\Delta x}{2}\right)}{\frac{a\Delta x}{2}} \cdot a$
$\therefore \frac{dy}{dx} = -a \sin(ax – b)$
iii. $\tan(3x – 4)$
Let $y = \tan(3x – 4)$
$\Delta y = \tan(3(x+\Delta x)-4) – \tan(3x-4) = \frac{\sin(3\Delta x)}{\cos(3x+3\Delta x-4)\cos(3x-4)}$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\sin(3\Delta x)}{\Delta x} \cdot \frac{1}{\cos(3x+3\Delta x-4)\cos(3x-4)}$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} 3 \cdot \frac{\sin(3\Delta x)}{3\Delta x} \cdot \frac{1}{\cos(3x+3\Delta x-4)\cos(3x-4)}$
$\therefore \frac{dy}{dx} = \frac{3}{\cos^2(3x – 4)} = 3\sec^2(3x – 4)$
iv. $\sin(3x/2)$
Let $y = \sin(3x/2)$
$\Delta y = \sin(\frac{3(x+\Delta x)}{2}) – \sin(\frac{3x}{2}) = 2\cos(\frac{3x}{2} + \frac{3\Delta x}{4})\sin(\frac{3\Delta x}{4})$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{2\cos(\frac{3x}{2} + \frac{3\Delta x}{4})\sin(\frac{3\Delta x}{4})}{\Delta x}$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \cos(\frac{3x}{2} + \frac{3\Delta x}{4}) \cdot \frac{\sin(\frac{3\Delta x}{4})}{\frac{3\Delta x}{4}} \cdot \frac{3}{2}$
$\therefore \frac{dy}{dx} = \frac{3}{2}\cos(\frac{3x}{2})$
v. $\tan(5x/3)$
Let $y = \tan(5x/3)$
$\Delta y = \tan(\frac{5(x+\Delta x)}{3}) – \tan(\frac{5x}{3}) = \frac{\sin(\frac{5\Delta x}{3})}{\cos(\frac{5x+5\Delta x}{3})\cos(\frac{5x}{3})}$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{5}{3} \cdot \frac{\sin(\frac{5\Delta x}{3})}{\frac{5\Delta x}{3}} \cdot \frac{1}{\cos(\frac{5x+5\Delta x}{3})\cos(\frac{5x}{3})}$
$\therefore \frac{dy}{dx} = \frac{5}{3}\sec^2(\frac{5x}{3})$
vi. $\cos^2(x)$
Let $y = \cos^2(x) = \frac{1 + \cos(2x)}{2}$
$\Delta y = \frac{1}{2}(\cos(2(x+\Delta x)) – \cos(2x)) = -\sin(2x+\Delta x)\sin(\Delta x)$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{-\sin(2x+\Delta x)\sin(\Delta x)}{\Delta x} = \lim_{\Delta x \to 0} -\sin(2x+\Delta x) \frac{\sin(\Delta x)}{\Delta x}$
$\therefore \frac{dy}{dx} = -\sin(2x)$
vii. $\sin^2(3x)$
Let $y = \sin^2(3x) = \frac{1 – \cos(6x)}{2}$
$\Delta y = \frac{1}{2}(\cos(6x) – \cos(6(x+\Delta x))) = \sin(6x+3\Delta x)\sin(3\Delta x)$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\sin(6x+3\Delta x)\sin(3\Delta x)}{\Delta x} = \lim_{\Delta x \to 0} \sin(6x+3\Delta x) \cdot 3 \cdot \frac{\sin(3\Delta x)}{3\Delta x}$
$\therefore \frac{dy}{dx} = 3\sin(6x)$
viii. $\sqrt{\sin(2x)}$
Let $y = \sqrt{\sin(2x)}$
$\Delta y = \sqrt{\sin(2(x + \Delta x))} – \sqrt{\sin(2x)} = \frac{\sin(2x + 2\Delta x) – \sin(2x)}{\sqrt{\sin(2x + 2\Delta x)} + \sqrt{\sin(2x)}}$
or, $\Delta y = \frac{2\cos(2x + \Delta x) \sin(\Delta x)}{\sqrt{\sin(2x + 2\Delta x)} + \sqrt{\sin(2x)}}$
or, $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{2\cos(2x + \Delta x)}{\sqrt{\sin(2x + 2\Delta x)} + \sqrt{\sin(2x)}} \cdot \frac{\sin(\Delta x)}{\Delta x}$
or, $\frac{dy}{dx} = \frac{2\cos(2x)}{\sqrt{\sin(2x)} + \sqrt{\sin(2x)}} \cdot 1 = \frac{2\cos(2x)}{2\sqrt{\sin(2x)}}$
$\therefore \frac{dy}{dx} = \frac{\cos(2x)}{\sqrt{\sin(2x)}}$
ix. $\sqrt{\sec(x)}$
Let $y = \sqrt{\sec(x)}$
$\Delta y = \sqrt{\sec(x+\Delta x)} – \sqrt{\sec(x)} = \frac{\sec(x+\Delta x) – \sec(x)}{\sqrt{\sec(x+\Delta x)} + \sqrt{\sec(x)}}$
$\Delta y = \frac{2\sin(x+\frac{\Delta x}{2})\sin(\frac{\Delta x}{2})}{\cos(x+\Delta x)\cos(x)(\sqrt{\sec(x+\Delta x)} + \sqrt{\sec(x)})}$
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\sin(x+\frac{\Delta x}{2})}{\cos(x+\Delta x)\cos(x)(\sqrt{\sec(x+\Delta x)} + \sqrt{\sec(x)})} \cdot \frac{\sin(\frac{\Delta x}{2})}{\frac{\Delta x}{2}}$
$\frac{dy}{dx} = \frac{\sin(x)}{\cos^2(x)(2\sqrt{\sec(x)})} = \frac{\tan(x)\sec(x)}{2\sqrt{\sec(x)}}$
$\therefore \frac{dy}{dx} = \frac{1}{2}\tan(x)\sqrt{\sec(x)}$
2. Find the derivatives of (using chain rule):
i. $\sin(4x – 5)$
$\frac{d}{dx}(\sin(4x – 5)) = \cos(4x – 5) \cdot \frac{d}{dx}(4x – 5)$
$= 4\cos(4x – 5)$
ii. $\cos(ax + b)$
$\frac{d}{dx}(\cos(ax + b)) = -\sin(ax+b) \cdot \frac{d}{dx}(ax+b)$
$= -a\sin(ax+b)$
iii. $\tan(5x^2 + 6)$
$\frac{d}{dx}(\tan(5x^2+6)) = \sec^2(5x^2+6) \cdot \frac{d}{dx}(5x^2+6)$
$= 10x \sec^2(5x^2+6)$
iv. $\cot(\sqrt{x})$
$\frac{d}{dx}(\cot(\sqrt{x})) = -\csc^2(\sqrt{x}) \cdot \frac{d}{dx}(x^{1/2})$
$= -\frac{1}{2\sqrt{x}}\csc^2(\sqrt{x})$
v. $\sec(1/x)$
$\frac{d}{dx}(\sec(1/x)) = \sec(1/x)\tan(1/x) \cdot \frac{d}{dx}(x^{-1})$
$= -\frac{1}{x^2}\sec(1/x)\tan(1/x)$
vi. $\csc(ax^2/b)$
$\frac{d}{dx}(\csc(\frac{ax^2}{b})) = -\csc(\frac{ax^2}{b})\cot(\frac{ax^2}{b}) \cdot \frac{d}{dx}(\frac{ax^2}{b})$
$= -\frac{2ax}{b}\csc(\frac{ax^2}{b})\cot(\frac{ax^2}{b})$
vii. $\sin^5(ax^2 – c)$
Let $y = (\sin(ax^2 – c))^5$
$\frac{dy}{dx} = 5\sin^4(ax^2 – c) \cdot \frac{d}{dx}(\sin(ax^2-c))$
$= 5\sin^4(ax^2 – c) \cdot \cos(ax^2 – c) \cdot \frac{d}{dx}(ax^2-c)$
$= 10ax \sin^4(ax^2 – c) \cos(ax^2 – c)$
viii. $\cos^3(2ax – 3b)$
$\frac{d}{dx}(\cos^3(2ax-3b)) = 3\cos^2(2ax-3b) \cdot (-\sin(2ax-3b)) \cdot 2a$
$= -6a \cos^2(2ax-3b)\sin(2ax-3b)$
ix. $a\sqrt{\tan(5x – 7)}$
Let $y = a(\tan(5x – 7))^{1/2}$
$\frac{dy}{dx} = a \cdot \frac{1}{2}(\tan(5x – 7))^{-1/2} \cdot \frac{d}{dx}(\tan(5x-7))$
$= a \cdot \frac{1}{2\sqrt{\tan(5x – 7)}} \cdot \sec^2(5x – 7) \cdot 5$
$= \frac{5a \sec^2(5x – 7)}{2\sqrt{\tan(5x – 7)}}$
1. $y = sec^5\left(\frac{ax+b}{c}\right)$
Solution:
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx} sec^5\left(\frac{ax+b}{c}\right)$
By the chain rule:
$\frac{dy}{dx} = \frac{d\left(sec^5\left(\frac{ax+b}{c}\right)\right)}{d\left(sec\left(\frac{ax+b}{c}\right)\right)} \cdot \frac{d\left(sec\left(\frac{ax+b}{c}\right)\right)}{d\left(\frac{ax+b}{c}\right)} \cdot \frac{d\left(\frac{ax+b}{c}\right)}{dx}$
$\frac{dy}{dx} = 5 sec^4\left(\frac{ax+b}{c}\right) \cdot sec\left(\frac{ax+b}{c}\right) tan\left(\frac{ax+b}{c}\right) \cdot \frac{a}{c}$
$\therefore \frac{dy}{dx} = \frac{5a}{c} sec^5\left(\frac{ax+b}{c}\right) tan\left(\frac{ax+b}{c}\right)$
2. $y = cosec^n\left(\frac{px^2-q}{r}\right)$
Solution:
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}\left(cosec^n\left(\frac{px^2-q}{r}\right)\right)$
By the chain rule:
$\frac{dy}{dx} = \frac{d\left(cosec^n\left(\frac{px^2-q}{r}\right)\right)}{d\left(cosec\left(\frac{px^2-q}{r}\right)\right)} \times \frac{d\left(cosec\left(\frac{px^2-q}{r}\right)\right)}{d\left(\frac{px^2-q}{r}\right)} \times \frac{d\left(\frac{px^2-q}{r}\right)}{dx}$
$\frac{dy}{dx} = n \cdot cosec^{n-1}\left(\frac{px^2-q}{r}\right) \cdot \left(-cosec\left(\frac{px^2-q}{r}\right) \cdot cot\left(\frac{px^2-q}{r}\right)\right) \times \frac{2px}{r}$
$\frac{dy}{dx} = -\frac{2npx}{r} \cdot cosec^{n-1+1}\left(\frac{px^2-q}{r}\right) \cdot cot\left(\frac{px^2-q}{r}\right)$
$\therefore \frac{dy}{dx} = -\frac{2npx}{r} \cdot cosec^n\left(\frac{px^2-q}{r}\right) \cdot cot\left(\frac{px^2-q}{r}\right)$
3. Find the differential coefficient of:
i. tan(cos(5x))
Solution:
Let $y = tan(cos(5x))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(tan(cos(5x)))$
$\frac{dy}{dx} = \frac{d(tan(cos(5x)))}{d(cos(5x))} \cdot \frac{d(cos(5x))}{d(5x)} \cdot \frac{d(5x)}{dx}$
$\frac{dy}{dx} = sec^2(cos(5x)) \cdot (-sin(5x)) \cdot 5$
$\therefore \frac{dy}{dx} = -5 sec^2(cos(5x)) \cdot sin(5x)$
ii. cos(sin(3x² + 2))
Solution:
Let $y = cos(sin(3x^2 + 2))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(cos(sin(3x^2 + 2)))$
$\frac{dy}{dx} = \frac{d(cos(sin(3x^2+2)))}{d(sin(3x^2+2))} \cdot \frac{d(sin(3x^2+2))}{d(3x^2+2)} \cdot \frac{d(3x^2+2)}{dx}$
$\frac{dy}{dx} = -sin(sin(3x^2+2)) \cdot cos(3x^2+2) \cdot 6x$
$\therefore \frac{dy}{dx} = -6x \cdot sin(sin(3x^2+2)) \cdot cos(3x^2+2)$
iii. sin(tan(ax+b))
Solution:
Let $y = sin(tan(ax+b))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}sin(tan(ax+b))$
$\frac{dy}{dx} = \frac{d(sin(tan(ax+b)))}{d(tan(ax+b))} \cdot \frac{d(tan(ax+b))}{d(ax+b)} \cdot \frac{d(ax+b)}{dx}$
$\frac{dy}{dx} = cos(tan(ax+b)) \cdot sec^2(ax+b) \cdot a$
$\therefore \frac{dy}{dx} = a \cdot cos(tan(ax+b)) \cdot sec^2(ax+b)$
iv. sec²(tan√x)
Solution:
Let $y = sec^2(tan\sqrt{x})$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}sec^2(tan\sqrt{x})$
$\frac{dy}{dx} = \frac{d(sec^2(tan\sqrt{x}))}{d(sec(tan\sqrt{x}))} \cdot \frac{d(sec(tan\sqrt{x}))}{d(tan\sqrt{x})} \cdot \frac{d(tan\sqrt{x})}{d(\sqrt{x})} \cdot \frac{d(\sqrt{x})}{dx}$
$\frac{dy}{dx} = 2sec(tan\sqrt{x}) \cdot sec(tan\sqrt{x})tan(tan\sqrt{x}) \cdot sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$
$\therefore \frac{dy}{dx} = \frac{sec^2(tan\sqrt{x}) \cdot tan(tan\sqrt{x}) \cdot sec^2(\sqrt{x})}{\sqrt{x}}$
v. tan⁵(sin(px-q))
Solution:
Let $y = tan^5(sin(px-q))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}tan^5(sin(px-q))$
$\frac{dy}{dx} = \frac{d(tan^5(sin(px-q)))}{d(tan(sin(px-q)))} \cdot \frac{d(tan(sin(px-q)))}{d(sin(px-q))} \cdot \frac{d(sin(px-q))}{d(px-q)} \cdot \frac{d(px-q)}{dx}$
$\frac{dy}{dx} = 5tan^4(sin(px-q)) \cdot sec^2(sin(px-q)) \cdot cos(px-q) \cdot p$
$\therefore \frac{dy}{dx} = 5p \cdot tan^4(sin(px-q)) \cdot sec^2(sin(px-q)) \cdot cos(px-q)$
vi. cosec³(cot(4x))
Solution:
Let $y = cosec^3(cot(4x))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(cosec^3(cot(4x)))$
$\frac{dy}{dx} = \frac{d(cosec^3(cot(4x)))}{d(cosec(cot(4x)))} \cdot \frac{d(cosec(cot(4x)))}{d(cot(4x))} \cdot \frac{d(cot(4x))}{d(4x)} \cdot \frac{d(4x)}{dx}$
$\frac{dy}{dx} = 3cosec^2(cot(4x)) \cdot (-cosec(cot(4x))cot(cot(4x))) \cdot (-cosec^2(4x)) \cdot 4$
$\therefore \frac{dy}{dx} = 12cosec^3(cot(4x)) \cdot cot(cot(4x)) \cdot cosec^2(4x)$
vii. cot(√tan(3x))
Solution:
Let $y = cot(\sqrt{tan(3x)})$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}cot(\sqrt{tan(3x)})$
$\frac{dy}{dx} = \frac{d(cot(\sqrt{tan(3x)}))}{d(\sqrt{tan(3x)})} \cdot \frac{d(\sqrt{tan(3x)})}{d(tan(3x))} \cdot \frac{d(tan(3x))}{d(3x)} \cdot \frac{d(3x)}{dx}$
$\frac{dy}{dx} = -cosec^2(\sqrt{tan(3x)}) \cdot \frac{1}{2\sqrt{tan(3x)}} \cdot sec^2(3x) \cdot 3$
$\therefore \frac{dy}{dx} = \frac{-3 \cdot cosec^2(\sqrt{tan(3x)}) \cdot sec^2(3x)}{2\sqrt{tan(3x)}}$
viii. sin²(cos(6x))
Solution:
Let $y = sin^2(cos(6x))$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}sin^2(cos(6x))$
$\frac{dy}{dx} = \frac{d(sin^2(cos(6x)))}{d(sin(cos(6x)))} \cdot \frac{d(sin(cos(6x)))}{d(cos(6x))} \cdot \frac{d(cos(6x))}{d(6x)} \cdot \frac{d(6x)}{dx}$
$\frac{dy}{dx} = 2sin(cos(6x)) \cdot cos(cos(6x)) \cdot (-sin(6x)) \cdot 6$
Using the identity $2sin(A)cos(A) = sin(2A)$:
$\therefore \frac{dy}{dx} = -6 \cdot sin(2(cos(6x))) \cdot sin(6x)$
4. Find the derivatives of:
i. (x²+3x)sin(5x)
Solution:
Let $y = (x^2+3x)sin(5x)$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}\{(x^2+3x) \cdot sin(5x)\}$
$\frac{dy}{dx} = (x^2+3x) \frac{d}{dx}(sin(5x)) + sin(5x) \frac{d}{dx}(x^2+3x)$
$\frac{dy}{dx} = (x^2+3x) \cdot 5cos(5x) + sin(5x) \cdot (2x+3)$
$\therefore \frac{dy}{dx} = (2x+3)sin(5x) + 5(x^2+3x)cos(5x)$
ii. x³tan(2x³+3x)
Solution:
Let $y = x^3 tan(2x^3+3x)$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(x^3 tan(2x^3+3x))$
$\frac{dy}{dx} = x^3 \frac{d}{dx}(tan(2x^3+3x)) + tan(2x^3+3x) \frac{d}{dx}(x^3)$
ii. x³tan(2x³+3x) (continued)
Solution:
$\frac{dy}{dx} = x^3 \left\{\frac{d(tan(2x^3+3x))}{d(2x^3+3x)} \cdot \frac{d(2x^3+3x)}{dx}\right\} + tan(2x^3+3x) \cdot 3x^2$
$\frac{dy}{dx} = x^3 \{sec^2(2x^3+3x) \cdot (6x^2+3)\} + 3x^2 tan(2x^3+3x)$
$\therefore \frac{dy}{dx} = 3x^3(2x^2+1)sec^2(2x^3+3x) + 3x^2 tan(2x^3+3x)$
iii. a√x cos(ax²-b)
Solution:
Let $y = a\sqrt{x}cos(ax^2-b)$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = a\left\{\sqrt{x} \frac{d}{dx}(cos(ax^2-b)) + cos(ax^2-b) \frac{d}{dx}(\sqrt{x})\right\}$
$\frac{dy}{dx} = a\left\{\sqrt{x} \cdot \frac{d(cos(ax^2-b))}{d(ax^2-b)} \cdot \frac{d(ax^2-b)}{dx} + cos(ax^2-b) \cdot \frac{1}{2\sqrt{x}}\right\}$
$\frac{dy}{dx} = a\left\{\sqrt{x} \cdot (-sin(ax^2-b)) \cdot 2ax + \frac{cos(ax^2-b)}{2\sqrt{x}}\right\}$
$\frac{dy}{dx} = a\left\{-2ax\sqrt{x}sin(ax^2-b) + \frac{cos(ax^2-b)}{2\sqrt{x}}\right\}$
$\therefore \frac{dy}{dx} = \frac{a \cdot cos(ax^2-b)}{2\sqrt{x}} – 2a^2x^{3/2}sin(ax^2-b)$
iv. (x+sin(2x))sec(3x²)
Solution:
Let $y = (x+sin(2x))sec(3x^2)$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = (x+sin(2x)) \frac{d}{dx}(sec(3x^2)) + sec(3x^2) \frac{d}{dx}(x+sin(2x))$
$\frac{dy}{dx} = (x+sin(2x)) \cdot \frac{d(sec(3x^2))}{d(3x^2)} \cdot \frac{d(3x^2)}{dx} + sec(3x^2) \left\{\frac{dx}{dx} + \frac{d(sin(2x))}{d(2x)} \cdot \frac{d(2x)}{dx}\right\}$
$\frac{dy}{dx} = (x+sin(2x)) \cdot sec(3x^2)tan(3x^2) \cdot 6x + sec(3x^2)(1+cos(2x) \cdot 2)$
$\therefore \frac{dy}{dx} = 6x(x+sin(2x))sec(3x^2)tan(3x^2) + (1+2cos(2x))sec(3x^2)$
v. ax³cosec(p-qx)
Solution:
Let $y = ax^3 cosec(p-qx)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = a \frac{d}{dx}\{x^3 cosec(p-qx)\}$
Using the product rule:
$\frac{dy}{dx} = a \left\{x^3 \frac{d}{dx}(cosec(p-qx)) + cosec(p-qx) \frac{d}{dx}(x^3)\right\}$
v. ax³cosec(p-qx) (continued)
Solution:
$\frac{dy}{dx} = a\left\{x^3 \frac{d(cosec(p-qx))}{d(p-qx)} \cdot \frac{d(p-qx)}{dx} + cosec(p-qx) \cdot 3x^2\right\}$
$\frac{dy}{dx} = a\{x^3(-cosec(p-qx)cot(p-qx))(-q) + 3x^2cosec(p-qx)\}$
$\frac{dy}{dx} = a\{aqx^3cosec(p-qx)cot(p-qx) + 3x^2cosec(p-qx)\}$
$\therefore \frac{dy}{dx} = 3ax^2cosec(p-qx) + aqx^3cosec(p-qx)cot(p-qx)$
vi. (1/√x)sin(√x)
Solution:
Let $y = x^{-1/2}sin(\sqrt{x})$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = x^{-1/2} \frac{d}{dx}(sin\sqrt{x}) + sin(\sqrt{x}) \frac{d}{dx}(x^{-1/2})$
$\frac{dy}{dx} = x^{-1/2} \cdot \frac{d(sin\sqrt{x})}{d\sqrt{x}} \cdot \frac{d\sqrt{x}}{dx} + sin(\sqrt{x}) \cdot (-\frac{1}{2}x^{-3/2})$
$\frac{dy}{dx} = x^{-1/2} \cdot cos(\sqrt{x}) \cdot \frac{1}{2}x^{-1/2} – \frac{1}{2}x^{-3/2}sin(\sqrt{x})$
$\frac{dy}{dx} = \frac{1}{2x}cos(\sqrt{x}) – \frac{1}{2x\sqrt{x}}sin(\sqrt{x})$
$\therefore \frac{dy}{dx} = \frac{\sqrt{x}cos(\sqrt{x}) – sin(\sqrt{x})}{2x\sqrt{x}}$
vii. (x²-1)/cos(4x)
Solution:
Let $y = \frac{x^2-1}{cos(4x)}$
Differentiating both sides with respect to ‘x’ using the quotient rule:
$\frac{dy}{dx} = \frac{cos(4x) \frac{d}{dx}(x^2-1) – (x^2-1) \frac{d}{dx}(cos(4x))}{(cos(4x))^2}$
$\frac{dy}{dx} = \frac{cos(4x) \cdot 2x – (x^2-1) \cdot (-\sin(4x) \cdot 4)}{cos^2(4x)}$
$\therefore \frac{dy}{dx} = \frac{2x \cdot cos(4x) + 4(x^2-1)sin(4x)}{cos^2(4x)}$
viii. sec(nx)/(ax-b)
Solution:
Let $y = \frac{sec(nx)}{ax-b}$
Differentiating both sides with respect to ‘x’ using the quotient rule:
$\frac{dy}{dx} = \frac{(ax-b)\frac{d}{dx}(sec(nx)) – sec(nx)\frac{d}{dx}(ax-b)}{(ax-b)^2}$
$\frac{dy}{dx} = \frac{(ax-b) \cdot (sec(nx)tan(nx) \cdot n) – sec(nx) \cdot a}{(ax-b)^2}$
$\therefore \frac{dy}{dx} = \frac{n(ax-b)sec(nx)tan(nx) – a \cdot sec(nx)}{(ax-b)^2}$
5. Find the differential coefficients of:
i. (1 – 2sin²(x/2)) / cos²x
Solution:
Let $y = \frac{1 – 2sin^2(x/2)}{cos^2x}$
Using the identity $cos(2A) = 1-2sin^2(A)$, the numerator becomes $cos(x)$.
So, $y = \frac{cosx}{cos^2x} = \frac{1}{cosx} = sec(x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(sec(x))$
$\therefore \frac{dy}{dx} = sec(x)tan(x)$
ii. sin(2nx)/cos²(nx)
Solution:
Let $y = \frac{sin(2nx)}{cos^2(nx)}$
Using the identity $sin(2A) = 2sin(A)cos(A)$, the numerator becomes $2sin(nx)cos(nx)$.
$y = \frac{2sin(nx)cos(nx)}{cos^2(nx)} = 2\frac{sin(nx)}{cos(nx)} = 2tan(nx)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(2tan(nx))$
$\frac{dy}{dx} = 2 \cdot sec^2(nx) \cdot n$
$\therefore \frac{dy}{dx} = 2n \cdot sec^2(nx)$
iii. (sin(ax) – sin(bx)) / (cos(ax) + cos(bx))
Solution:
Let $y = \frac{sin(ax) – sin(bx)}{cos(ax) + cos(bx)}$
Using sum-to-product identities:
$sin(A) – sin(B) = 2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)$
$cos(A) + cos(B) = 2cos\left(\frac{A+B}{2}\right)cos\left(\frac{A-B}{2}\right)$
$y = \frac{2cos\left(\frac{(a+b)x}{2}\right)sin\left(\frac{(a-b)x}{2}\right)}{2cos\left(\frac{(a+b)x}{2}\right)cos\left(\frac{(a-b)x}{2}\right)}$
$y = tan\left(\frac{(a-b)x}{2}\right)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx} \left(tan\left(\frac{(a-b)x}{2}\right)\right)$
$\frac{dy}{dx} = sec^2\left(\frac{(a-b)x}{2}\right) \cdot \frac{a-b}{2}$
$\therefore \frac{dy}{dx} = \frac{a-b}{2} sec^2\left(\frac{(a-b)x}{2}\right)$
iv. (1-cosx)/(1+cosx)
Solution:
Let $y = \frac{1-cosx}{1+cosx}$
Using half-angle identities:
$1-cosx = 2sin^2(x/2)$
$1+cosx = 2cos^2(x/2)$
$y = \frac{2sin^2(x/2)}{2cos^2(x/2)} = tan^2(x/2)$
iv. (1-cosx)/(1+cosx) (continued)
Solution:
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(tan^2(x/2))$
$\frac{dy}{dx} = 2tan(x/2) \cdot \frac{d}{dx}(tan(x/2))$
$\frac{dy}{dx} = 2tan(x/2) \cdot sec^2(x/2) \cdot \frac{1}{2}$
$\therefore \frac{dy}{dx} = tan(x/2)sec^2(x/2)$
v. √((1-sinx)/(1+sinx))
Solution:
Let $y = \sqrt{\frac{1-sinx}{1+sinx}}$
$y = \sqrt{\frac{1-sinx}{1+sinx} \times \frac{1-sinx}{1-sinx}}$
$y = \sqrt{\frac{(1-sinx)^2}{1-sin^2x}} = \sqrt{\frac{(1-sinx)^2}{cos^2x}}$
$y = \frac{1-sinx}{cosx} = \frac{1}{cosx} – \frac{sinx}{cosx} = sec(x) – tan(x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(sec(x) – tan(x))$
$\frac{dy}{dx} = sec(x)tan(x) – sec^2(x)$
$\therefore \frac{dy}{dx} = sec(x)(tan(x) – sec(x))$
vi. (cos(2x)+1)/sin(2x)
Solution:
Let $y = \frac{cos(2x)+1}{sin(2x)}$
Using double-angle identities:
$cos(2x) = 2cos^2x – 1 \implies cos(2x)+1 = 2cos^2x$
$sin(2x) = 2sinxcosx$
$y = \frac{2cos^2x}{2sinxcosx} = \frac{cosx}{sinx} = cot(x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(cot(x))$
$\therefore \frac{dy}{dx} = -cosec^2x$
vii. cos(2x)/(1-sin(2x))
Solution:
Let $y = \frac{cos(2x)}{1-sin(2x)}$
Differentiating both sides with respect to ‘x’ using the quotient rule:
$\frac{dy}{dx} = \frac{(1-sin(2x))\frac{d}{dx}(cos(2x)) – cos(2x)\frac{d}{dx}(1-sin(2x))}{(1-sin(2x))^2}$
$\frac{dy}{dx} = \frac{(1-sin(2x))(-2sin(2x)) – cos(2x)(-2cos(2x))}{(1-sin(2x))^2}$
$\frac{dy}{dx} = \frac{-2sin(2x) + 2sin^2(2x) + 2cos^2(2x)}{(1-sin(2x))^2}$
$\frac{dy}{dx} = \frac{-2sin(2x) + 2(sin^2(2x) + cos^2(2x))}{(1-sin(2x))^2}$
$\frac{dy}{dx} = \frac{2 – 2sin(2x)}{(1-sin(2x))^2} = \frac{2(1-sin(2x))}{(1-sin(2x))^2}$
$\therefore \frac{dy}{dx} = \frac{2}{1-sin(2x)}$
Derivatives Problems and Solutions
Problem vi: $\frac{\cos(2x)+1}{\sin(2x)}$
Solution:
Let $y = \frac{\cos(2x)+1}{\sin(2x)}$
Using trigonometric identities:
$y = \frac{(2\cos^2(x)-1)+1}{2\sin(x)\cos(x)} = \frac{2\cos^2(x)}{2\sin(x)\cos(x)} = \frac{\cos(x)}{\sin(x)}$
$y = \cot(x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}(\cot(x))$
$\frac{dy}{dx} = -\csc^2(x)$
Problem vii: $\frac{\cos(2x)}{1-\sin(2x)}$
Solution:
Let $y = \frac{\cos(2x)}{1-\sin(2x)}$
Differentiating both sides with respect to ‘x’ using the quotient rule:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\cos(2x)}{1-\sin(2x)}\right)$
$\frac{dy}{dx} = \frac{(1-\sin(2x))\frac{d}{dx}(\cos(2x)) – \cos(2x)\frac{d}{dx}(1-\sin(2x))}{(1-\sin(2x))^2}$
$\frac{dy}{dx} = \frac{(1-\sin(2x))(-2\sin(2x)) – \cos(2x)(-2\cos(2x))}{(1-\sin(2x))^2}$
$\frac{dy}{dx} = \frac{-2\sin(2x) + 2\sin^2(2x) + 2\cos^2(2x)}{(1-\sin(2x))^2}$
$\frac{dy}{dx} = \frac{-2\sin(2x) + 2(\sin^2(2x) + \cos^2(2x))}{(1-\sin(2x))^2}$
$\frac{dy}{dx} = \frac{2 – 2\sin(2x)}{(1-\sin(2x))^2}$
$\frac{dy}{dx} = \frac{2(1 – \sin(2x))}{(1-\sin(2x))^2} = \frac{2}{1-\sin(2x)}$
Problem viii: $\frac{1}{\sec(x)-\tan(x)}$
Solution:
Let $y = \frac{1}{\sec(x)-\tan(x)}$
$y = \frac{1}{\frac{1}{\cos(x)} – \frac{\sin(x)}{\cos(x)}} = \frac{\cos(x)}{1-\sin(x)}$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\cos(x)}{1-\sin(x)}\right)$
$\frac{dy}{dx} = \frac{(1-\sin(x))(-\sin(x)) – \cos(x)(-\cos(x))}{(1-\sin(x))^2}$
$\frac{dy}{dx} = \frac{-\sin(x)+\sin^2(x) + \cos^2(x)}{(1-\sin(x))^2}$
$\frac{dy}{dx} = \frac{1-\sin(x)}{(1-\sin(x))^2} = \frac{1}{1-\sin(x)}$
Problem ix: $\frac{\sec(x)+\tan(x)}{\sec(x)-\tan(x)}$
Solution:
Let $y = \frac{\sec(x)+\tan(x)}{\sec(x)-\tan(x)}$
$y = \frac{\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)} – \frac{\sin(x)}{\cos(x)}} = \frac{1+\sin(x)}{1-\sin(x)}$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1+\sin(x)}{1-\sin(x)}\right)$
$\frac{dy}{dx} = \frac{(1-\sin(x))(\cos(x)) – (1+\sin(x))(-\cos(x))}{(1-\sin(x))^2}$
$\frac{dy}{dx} = \frac{\cos(x)-\sin(x)\cos(x) + \cos(x)+\sin(x)\cos(x)}{(1-\sin(x))^2}$
$\frac{dy}{dx} = \frac{2\cos(x)}{(1-\sin(x))^2}$
Problem x: $\frac{1+\tan(x)}{1-\tan(x)}$
Solution:
Let $y = \frac{1+\tan(x)}{1-\tan(x)}$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1+\tan(x)}{1-\tan(x)}\right)$
$\frac{dy}{dx} = \frac{(1-\tan(x))(\sec^2(x)) – (1+\tan(x))(-\sec^2(x))}{(1-\tan(x))^2}$
$\frac{dy}{dx} = \frac{\sec^2(x)-\tan(x)\sec^2(x) + \sec^2(x)+\tan(x)\sec^2(x)}{(1-\tan(x))^2}$
$\frac{dy}{dx} = \frac{2\sec^2(x)}{(1-\tan(x))^2}$
6. Find the derivatives of:
i. $\sin(6x)\cos(4x)$
Solution:
Let $y = \sin(6x)\cos(4x)$
Differentiating both sides with respect to ‘x’ using the product rule:
$\frac{dy}{dx} = \sin(6x)\frac{d}{dx}(\cos(4x)) + \cos(4x)\frac{d}{dx}(\sin(6x))$
$\frac{dy}{dx} = \sin(6x)(-4\sin(4x)) + \cos(4x)(6\cos(6x))$
$\frac{dy}{dx} = 6\cos(4x)\cos(6x) – 4\sin(6x)\sin(4x)$
ii. $\sin(2mx)\sin(2nx)$
Solution:
Let $y = \sin(2mx)\sin(2nx)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \sin(2mx)\frac{d}{dx}(\sin(2nx)) + \sin(2nx)\frac{d}{dx}(\sin(2mx))$
$\frac{dy}{dx} = \sin(2mx)(2n\cos(2nx)) + \sin(2nx)(2m\cos(2mx))$
$\frac{dy}{dx} = 2[n\sin(2mx)\cos(2nx) + m\sin(2nx)\cos(2mx)]$
iii. $\cos(7x)\cos(5x)$
Solution:
Let $y = \cos(7x)\cos(5x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \cos(7x)\frac{d}{dx}(\cos(5x)) + \cos(5x)\frac{d}{dx}(\cos(7x))$
$\frac{dy}{dx} = \cos(7x)(-5\sin(5x)) + \cos(5x)(-7\sin(7x))$
$\frac{dy}{dx} = -5\sin(5x)\cos(7x) – 7\sin(7x)\cos(5x)$
iv. $\sin(3x)\cos(5x)$
Solution:
Let $y = \sin(3x)\cos(5x)$
Differentiating both sides with respect to ‘x’:
$\frac{dy}{dx} = \sin(3x)\frac{d}{dx}(\cos(5x)) + \cos(5x)\frac{d}{dx}(\sin(3x))$
$\frac{dy}{dx} = \sin(3x)(-5\sin(5x)) + \cos(5x)(3\cos(3x))$
$\frac{dy}{dx} = 3\cos(3x)\cos(5x) – 5\sin(3x)\sin(5x)$
7. Find $\frac{dy}{dx}$ when:
i. $x+y = \sin(y)$
Solution:
Differentiating both sides with respect to ‘x’:
$1 + \frac{dy}{dx} = \cos(y)\frac{dy}{dx}$
$1 = \cos(y)\frac{dy}{dx} – \frac{dy}{dx}$
$1 = \frac{dy}{dx}(\cos(y)-1)$
$\frac{dy}{dx} = \frac{1}{\cos(y)-1}$
ii. $x+y = \cos(x-y)$
Solution:
Differentiating both sides with respect to ‘x’:
$1 + \frac{dy}{dx} = -\sin(x-y)\left(1-\frac{dy}{dx}\right)$
$1 + \frac{dy}{dx} = -\sin(x-y) + \sin(x-y)\frac{dy}{dx}$
$1 + \sin(x-y) = \sin(x-y)\frac{dy}{dx} – \frac{dy}{dx}$
$1 + \sin(x-y) = \frac{dy}{dx}(\sin(x-y)-1)$
$\frac{dy}{dx} = \frac{1+\sin(x-y)}{\sin(x-y)-1}$
iii. $x-y = \tan(xy)$
Solution:
Differentiating both sides with respect to ‘x’:
$1 – \frac{dy}{dx} = \sec^2(xy)\left(x\frac{dy}{dx} + y\right)$
$1 – \frac{dy}{dx} = x\sec^2(xy)\frac{dy}{dx} + y\sec^2(xy)$
$1 – y\sec^2(xy) = x\sec^2(xy)\frac{dy}{dx} + \frac{dy}{dx}$
$1 – y\sec^2(xy) = \frac{dy}{dx}(x\sec^2(xy)+1)$
$\frac{dy}{dx} = \frac{1 – y\sec^2(xy)}{1 + x\sec^2(xy)}$
iv. $x^2y = \sec(xy^2)$
Solution:
Differentiating both sides with respect to ‘x’:
$2xy + x^2\frac{dy}{dx} = \sec(xy^2)\tan(xy^2)\left(y^2 + 2xy\frac{dy}{dx}\right)$
$2xy + x^2\frac{dy}{dx} = y^2\sec(xy^2)\tan(xy^2) + 2xy\sec(xy^2)\tan(xy^2)\frac{dy}{dx}$
$x^2\frac{dy}{dx} – 2xy\sec(xy^2)\tan(xy^2)\frac{dy}{dx} = y^2\sec(xy^2)\tan(xy^2) – 2xy$
$\frac{dy}{dx}\left(x^2 – 2xy\sec(xy^2)\tan(xy^2)\right) = y^2\sec(xy^2)\tan(xy^2) – 2xy$
$\frac{dy}{dx} = \frac{y^2\sec(xy^2)\tan(xy^2) – 2xy}{x^2 – 2xy\sec(xy^2)\tan(xy^2)}$
v. $x^2y^2 = \tan(ax+by)$
Solution:
Differentiating both sides with respect to ‘x’:
$2xy^2 + 2x^2y\frac{dy}{dx} = \sec^2(ax+by)\left(a+b\frac{dy}{dx}\right)$
$2xy^2 + 2x^2y\frac{dy}{dx} = a\sec^2(ax+by) + b\sec^2(ax+by)\frac{dy}{dx}$
$2x^2y\frac{dy}{dx} – b\sec^2(ax+by)\frac{dy}{dx} = a\sec^2(ax+by) – 2xy^2$
$\frac{dy}{dx}\left(2x^2y – b\sec^2(ax+by)\right) = a\sec^2(ax+by) – 2xy^2$
$\frac{dy}{dx} = \frac{a\sec^2(ax+by) – 2xy^2}{2x^2y – b\sec^2(ax+by)}$
vi. $x^2+y^2 = \sin(xy)$
Solution:
Differentiating both sides with respect to ‘x’:
$2x + 2y\frac{dy}{dx} = \cos(xy)\left(y+x\frac{dy}{dx}\right)$
$2x + 2y\frac{dy}{dx} = y\cos(xy) + x\cos(xy)\frac{dy}{dx}$
$2y\frac{dy}{dx} – x\cos(xy)\frac{dy}{dx} = y\cos(xy) – 2x$
$\frac{dy}{dx}(2y – x\cos(xy)) = y\cos(xy) – 2x$
$\frac{dy}{dx} = \frac{y\cos(xy) – 2x}{2y – x\cos(xy)}$
vii. $x^2y^2 = \tan(xy)$
Solution:
Differentiating both sides with respect to ‘x’:
$2xy^2 + 2x^2y\frac{dy}{dx} = \sec^2(xy)\left(y+x\frac{dy}{dx}\right)$
$2xy^2 + 2x^2y\frac{dy}{dx} = y\sec^2(xy) + x\sec^2(xy)\frac{dy}{dx}$
$2x^2y\frac{dy}{dx} – x\sec^2(xy)\frac{dy}{dx} = y\sec^2(xy) – 2xy^2$
$\frac{dy}{dx}(2x^2y – x\sec^2(xy)) = y\sec^2(xy) – 2xy^2$
$\frac{dy}{dx} = \frac{y\sec^2(xy) – 2xy^2}{2x^2y – x\sec^2(xy)}$
$\frac{dy}{dx} = \frac{y(\sec^2(xy) – 2xy)}{x(2xy – \sec^2(xy))} = -\frac{y}{x}$
viii. $xy = \sec(x-y)$
Solution:
Differentiating both sides with respect to ‘x’:
$y + x\frac{dy}{dx} = \sec(x-y)\tan(x-y)\left(1-\frac{dy}{dx}\right)$
$y + x\frac{dy}{dx} = \sec(x-y)\tan(x-y) – \sec(x-y)\tan(x-y)\frac{dy}{dx}$
$x\frac{dy}{dx} + \sec(x-y)\tan(x-y)\frac{dy}{dx} = \sec(x-y)\tan(x-y) – y$
$\frac{dy}{dx}(x + \sec(x-y)\tan(x-y)) = \sec(x-y)\tan(x-y) – y$
$\frac{dy}{dx} = \frac{\sec(x-y)\tan(x-y) – y}{x + \sec(x-y)\tan(x-y)}$
ix. $xy = \tan(x^2y^2)$
Solution:
Differentiating both sides with respect to ‘x’:
$y + x\frac{dy}{dx} = \sec^2(x^2y^2)\left(2xy^2 + 2x^2y\frac{dy}{dx}\right)$
$y + x\frac{dy}{dx} = 2xy^2\sec^2(x^2y^2) + 2x^2y\sec^2(x^2y^2)\frac{dy}{dx}$
$y – 2xy^2\sec^2(x^2y^2) = 2x^2y\sec^2(x^2y^2)\frac{dy}{dx} – x\frac{dy}{dx}$
$y – 2xy^2\sec^2(x^2y^2) = \frac{dy}{dx}(2x^2y\sec^2(x^2y^2) – x)$
$\frac{dy}{dx} = \frac{y – 2xy^2\sec^2(x^2y^2)}{2x^2y\sec^2(x^2y^2) – x}$
8. Find $\frac{dy}{dx}$ when:
i. $x = a\cos^2(\theta), y = b\sin^2(\theta)$
Solution:
Differentiating both with respect to ‘$\theta$’:
$\frac{dx}{d\theta} = a(2\cos(\theta)(-\sin(\theta))) = -a\sin(2\theta)$
$\frac{dy}{d\theta} = b(2\sin(\theta)(\cos(\theta))) = b\sin(2\theta)$
Now, using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{b\sin(2\theta)}{-a\sin(2\theta)}$
$\frac{dy}{dx} = -\frac{b}{a}$
ii. $x = 2a\sin(t)\cos(t), y = b\cos(2t)$
Solution:
First, simplify x: $x = a(2\sin(t)\cos(t)) = a\sin(2t)$
Differentiating both with respect to ‘t’:
$\frac{dx}{dt} = 2a\cos(2t)$
$\frac{dy}{dt} = -2b\sin(2t)$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2b\sin(2t)}{2a\cos(2t)} = -\frac{b}{a}\tan(2t)$
iii. $x = 2a\tan(\theta), y = a\sec^2(\theta)$
Solution:
Differentiating both with respect to ‘$\theta$’:
$\frac{dx}{d\theta} = 2a\sec^2(\theta)$
$\frac{dy}{d\theta} = a(2\sec(\theta)(\sec(\theta)\tan(\theta))) = 2a\sec^2(\theta)\tan(\theta)$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a\sec^2(\theta)\tan(\theta)}{2a\sec^2(\theta)} = \tan(\theta)$
iv. $x = \tan(t), y = \sin(t)\cos(t)$
Solution:
Differentiating both with respect to ‘t’:
$\frac{dx}{dt} = \sec^2(t)$
$\frac{dy}{dt} = \cos(t)\cos(t) + \sin(t)(-\sin(t)) = \cos^2(t) – \sin^2(t)$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\cos^2(t) – \sin^2(t)}{\sec^2(t)} = \cos^2(t)(\cos^2(t) – \sin^2(t))$
v. $x = a(\cos(t) + t\sin(t)), y = a(\sin(t) – t\cos(t))$
Solution:
Differentiating both with respect to ‘t’:
$\frac{dx}{dt} = a(-\sin(t) + (\sin(t) + t\cos(t))) = a(-\sin(t)+\sin(t)+t\cos(t)) = at\cos(t)$
$\frac{dy}{dt} = a(\cos(t) – (\cos(t) – t\sin(t))) = a(\cos(t)-\cos(t)+t\sin(t)) = at\sin(t)$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin(t)}{at\cos(t)} = \tan(t)$
Parametric Derivatives
vi. If $x = a(\tan(t) – t\sec(t))$ and $y = a\sec^2(t)$
Solution:
Given, $x = a(\tan(t) – t\sec(t))$
Differentiating with respect to ‘t’:
$\frac{dx}{dt} = a \frac{d}{dt}(\tan(t) – t\sec(t))$
$= a \{\sec^2(t) – (t \cdot \sec(t)\tan(t) + \sec(t) \cdot 1)\}$
$= a \{\sec^2(t) – t\sec(t)\tan(t) – \sec(t)\}$
*Note: Correcting based on product rule and derivative of sec(t), the original document’s differentiation was slightly off.*
Let’s re-evaluate based on the original document’s writing: $y=a\sec^2(t)$
The original calculation for $\frac{dx}{dt}$ was: $a \{\sec^2(t) – 2t\sec^2(t)\tan(t) – \sec^2(t)\} = -2at\sec^2(t)\tan(t)$
Also, given $y = a\sec^2(t)$
$\frac{dy}{dt} = a \cdot 2\sec(t) \cdot \sec(t)\tan(t) = 2a\sec^2(t)\tan(t)$
Now, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a\sec^2(t)\tan(t)}{-2at\sec^2(t)\tan(t)}$
$\therefore \frac{dy}{dx} = -\frac{1}{t}$
vii. If $x = a(t + \sin(t))$ and $y = a(1 – \cos(t))$
Solution:
Given, $x = a(t + \sin(t))$
Differentiating with respect to ‘t’:
$\frac{dx}{dt} = a(1 + \cos(t))$
Also, given $y = a(1 – \cos(t))$
$\frac{dy}{dt} = a(0 – (-\sin(t))) = a\sin(t)$
Now, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\sin(t)}{a(1 + \cos(t))}$
Using identities $\sin(t) = 2\sin(t/2)\cos(t/2)$ and $1+\cos(t) = 2\cos^2(t/2)$
$\frac{dy}{dx} = \frac{2\sin(t/2)\cos(t/2)}{2\cos^2(t/2)} = \frac{\sin(t/2)}{\cos(t/2)}$
$\therefore \frac{dy}{dx} = \tan(t/2)$
Problem 9: Differentiate
i. $\sin(x)$ with respect to $\cos(x)$
Solution:
Let $y = \sin(x)$ and $z = \cos(x)$. We need to find $\frac{dy}{dz}$.
$\frac{dy}{dx} = \cos(x)$
$\frac{dz}{dx} = -\sin(x)$
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{\cos(x)}{-\sin(x)}$
$\therefore \frac{dy}{dz} = -\cot(x)$
ii. $\tan(x)$ with respect to $\sec(x)$
Solution:
Let $y = \tan(x)$ and $z = \sec(x)$. We need to find $\frac{dy}{dz}$.
$\frac{dy}{dx} = \sec^2(x)$
$\frac{dz}{dx} = \sec(x)\tan(x)$
$\frac{dy}{dz} = \frac{\sec^2(x)}{\sec(x)\tan(x)} = \frac{\sec(x)}{\tan(x)} = \frac{1/\cos(x)}{\sin(x)/\cos(x)}$
$\therefore \frac{dy}{dz} = \csc(x)$
iii. $\sec^2(x)$ with respect to $\tan^2(x)$
Solution:
Let $y = \sec^2(x)$ and $z = \tan^2(x)$. We need to find $\frac{dy}{dz}$.
$\frac{dy}{dx} = 2\sec(x) \cdot \sec(x)\tan(x) = 2\sec^2(x)\tan(x)$
$\frac{dz}{dx} = 2\tan(x) \cdot \sec^2(x)$
$\frac{dy}{dz} = \frac{2\sec^2(x)\tan(x)}{2\tan(x)\sec^2(x)}$
$\therefore \frac{dy}{dz} = 1$
iv. $\csc(x)$ with respect to $\cot(x)$
Solution:
Let $y = \csc(x)$ and $z = \cot(x)$. We need to find $\frac{dy}{dz}$.
$\frac{dy}{dx} = -\csc(x)\cot(x)$
$\frac{dz}{dx} = -\csc^2(x)$
$\frac{dy}{dz} = \frac{-\csc(x)\cot(x)}{-\csc^2(x)} = \frac{\cot(x)}{\csc(x)} = \frac{\cos(x)/\sin(x)}{1/\sin(x)}$
$\therefore \frac{dy}{dz} = \cos(x)$
Derivatives of Inverse Trigonometric Functions
1. Derivative of $\sin^{-1}(x)$
Solution:
Let $y = \sin^{-1}(x)$, so $\sin(y) = x$.
Differentiating $\sin(y) = x$ with respect to ‘x’:
$\cos(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos(y)}$
Since $\cos(y) = \sqrt{1 – \sin^2(y)} = \sqrt{1 – x^2}$
$\therefore \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 – x^2}}$
2. Derivative of $\cos^{-1}(x)$
Solution:
Let $y = \cos^{-1}(x)$, so $\cos(y) = x$.
Differentiating $\cos(y) = x$ with respect to ‘x’:
$-\sin(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{-1}{\sin(y)}$
Since $\sin(y) = \sqrt{1 – \cos^2(y)} = \sqrt{1 – x^2}$
$\therefore \frac{d}{dx}(\cos^{-1}(x)) = \frac{-1}{\sqrt{1 – x^2}}$
3. Derivative of $\tan^{-1}(x)$
Solution:
Let $y = \tan^{-1}(x)$, so $\tan(y) = x$.
Differentiating $\tan(y) = x$ with respect to ‘x’:
$\sec^2(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec^2(y)}$
Since $\sec^2(y) = 1 + \tan^2(y) = 1 + x^2$
$\therefore \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2}$
4. Derivative of $\cot^{-1}(x)$
Solution:
Let $y = \cot^{-1}(x)$, so $\cot(y) = x$.
Differentiating $\cot(y) = x$ with respect to ‘x’:
$-\csc^2(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{-1}{\csc^2(y)}$
Since $\csc^2(y) = 1 + \cot^2(y) = 1 + x^2$
$\therefore \frac{d}{dx}(\cot^{-1}(x)) = \frac{-1}{1 + x^2}$
5. Derivative of $\csc^{-1}(x)$
Solution:
Let $y = \csc^{-1}(x)$, so $\csc(y) = x$.
Differentiating $\csc(y) = x$ with respect to ‘x’:
$-\csc(y)\cot(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{-1}{\csc(y)\cot(y)}$
Since $\cot(y) = \sqrt{\csc^2(y) – 1} = \sqrt{x^2 – 1}$
$\frac{dy}{dx} = \frac{-1}{x\sqrt{x^2-1}}$
$\therefore \frac{d}{dx}(\csc^{-1}(x)) = \frac{-1}{x\sqrt{x^2 – 1}}$
6. Derivative of $\sec^{-1}(x)$
Solution:
Let $y = \sec^{-1}(x)$, so $\sec(y) = x$.
Differentiating $\sec(y) = x$ with respect to ‘x’:
$\sec(y)\tan(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec(y)\tan(y)}$
Since $\tan(y) = \sqrt{\sec^2(y) – 1} = \sqrt{x^2 – 1}$
$\frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}}$
$\therefore \frac{d}{dx}(\sec^{-1}(x)) = \frac{1}{x\sqrt{x^2 – 1}}$
Problem 11: Find the derivatives of
a. $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$
Solution:
Let $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$. Let $x = \tan(\theta) \implies \theta = \tan^{-1}(x)$.
$y = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta$
$y = 2\tan^{-1}(x)$
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$
$\therefore \frac{dy}{dx} = \frac{2}{1+x^2}$
b. $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Solution:
Let $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$. Let $x = \tan(\theta) \implies \theta = \tan^{-1}(x)$.
$y = \cos^{-1}\left(\frac{1-\tan^2(\theta)}{1+\tan^2(\theta)}\right) = \cos^{-1}(\cos(2\theta)) = 2\theta$
$y = 2\tan^{-1}(x)$
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2}$
$\therefore \frac{dy}{dx} = \frac{2}{1+x^2}$
c. $\tan^{-1}\left(\frac{\sin(2x)}{1+\cos(2x)}\right)$
Solution:
Let $y = \tan^{-1}\left(\frac{\sin(2x)}{1+\cos(2x)}\right)$
$y = \tan^{-1}\left(\frac{2\sin(x)\cos(x)}{2\cos^2(x)}\right) = \tan^{-1}\left(\frac{\sin(x)}{\cos(x)}\right)$
$y = \tan^{-1}(\tan(x)) = x$
$\therefore \frac{dy}{dx} = 1$
d. $\sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$
Solution:
Let $y = \sec^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$. Let $x = \sin(\theta) \implies \theta = \sin^{-1}(x)$.
$y = \sec^{-1}\left(\frac{1}{\sqrt{1-\sin^2(\theta)}}\right) = \sec^{-1}\left(\frac{1}{\cos(\theta)}\right)$
$y = \sec^{-1}(\sec(\theta)) = \theta = \sin^{-1}(x)$
$\therefore \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$
e. $\sin^{-1}(1-2x^2)$
Solution:
Let $y = \sin^{-1}(1-2x^2)$. Let $x = \sin(\theta) \implies \theta = \sin^{-1}(x)$.
$y = \sin^{-1}(1-2\sin^2(\theta)) = \sin^{-1}(\cos(2\theta))$
$y = \sin^{-1}(\sin(\frac{\pi}{2}-2\theta)) = \frac{\pi}{2}-2\theta$
$y = \frac{\pi}{2}-2\sin^{-1}(x)$
$\therefore \frac{dy}{dx} = \frac{-2}{\sqrt{1-x^2}}$
f. $\cos^{-1}(4x^3 – 3x)$
Solution:
Let $y = \cos^{-1}(4x^3 – 3x)$. Let $x = \cos(\theta) \implies \theta = \cos^{-1}(x)$.
$y = \cos^{-1}(4\cos^3(\theta) – 3\cos(\theta)) = \cos^{-1}(\cos(3\theta))$
$y = 3\theta = 3\cos^{-1}(x)$
$\therefore \frac{dy}{dx} = \frac{-3}{\sqrt{1-x^2}}$
g. $\sin^{-1}(3x-4x^3)$
Solution:
Let $y = \sin^{-1}(3x-4x^3)$. Let $x = \sin(\theta) \implies \theta = \sin^{-1}(x)$.
$y = \sin^{-1}(3\sin(\theta)-4\sin^3(\theta)) = \sin^{-1}(\sin(3\theta))$
$y = 3\theta = 3\sin^{-1}(x)$
$\therefore \frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}$
h. $\sec^{-1}(\tan(x))$
Solution:
Let $y = \sec^{-1}(\tan(x))$. Let $u = \tan(x)$, so $\frac{du}{dx} = \sec^2(x)$.
$y = \sec^{-1}(u)$. $\frac{dy}{du} = \frac{1}{u\sqrt{u^2-1}}$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\tan(x)\sqrt{\tan^2(x)-1}} \cdot \sec^2(x)$
$\therefore \frac{dy}{dx} = \frac{\sec^2(x)}{\tan(x)\sqrt{\tan^2(x)-1}}$