Question

find the derivative of cot(3x-2) with respect to x from First principle of derivatives.

Answer 1

Answer:

Step-by-step explanation:

First Derivative of a function f(x) can be found out by the rule

$\frac{dy}{dx}= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\\\ \frac{dy}{dx}= \lim_{h \to 0} \frac{cot [3 (x+h)-2]-cot (3 x-2)}{h}\\\\ \frac{dy}{dx}= \lim_{h \to 0}\frac{ \frac{cos[ 3 (x+h)-2]}{sin(3x+3h-2)}- \frac{cos(3 x-2)}{sin(3 x-2)}}{h}\\\\ \frac{dy}{dx}= \lim_{h \to 0} \frac{cos [3(x+h)-2]*sin (3 x-2) - sin [3(x+h)-2]*cos (3 x-2) }{h*sin(3x+3h-2)*sin(3x-2)}$

$\\\\ \frac{dy}{dx}= \lim_{h \to 0} \frac {sin(3x-2-3x-3h+2)}{sin^2(3x-2)*h}\\\\\frac{dy}{dx}= \lim_{h \to 0} \frac{[sin(-3 h)]}{-3h}*[-3*cosec^2(3x-2)]\\\\=-3cosec^2(3x-2)$

Used the following trigonometric as well as Theorem on limits

$\lim_{x \to 0} \frac{sinx}{x}=1\\\\ sin (A-B)=sin A *cos B - cos A* sin B$

f'[cot(3 x-2)]= -3 cosec²(3x-2)