Question

if y = x/1+cot x , find dy/dx​

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given That :- }}}}$

• y = x / 1 + cot x .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find:- }}}}$

• The derivative of the given function , i.e. dy/dx .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution:- }}}}$

A function is given to us and we need to find out its derivative . The given function is :-

$\sf\dashrightarrow y = \dfrac{ x }{1+cot(x)}$

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We can use the ❝ Quoteint Rule ❞ , as ,

$\dashrightarrow \boxed{\red{\sf \dfrac{d}{dx}\bigg[ \dfrac{ f(x)}{g(x)}\bigg] = \dfrac{ g(x)\bigg( \dfrac{d}{dx}[ f(x)] \bigg)- f(x)\bigg( \dfrac{d}{dx}[ g(x)]\bigg) }{ [ g(x)]^2} }}$

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❒ Differenciate both sides wrt x :-

$\sf\dashrightarrow\dfrac{dy}{dx}= \dfrac{d\bigg( \dfrac{x}{1+cot x} \bigg) }{dx}$

$\sf\dashrightarrow\dfrac{dy}{dx}= \dfrac{ (1+cot\ x)\dfrac{dx}{dx} - x \bigg( \dfrac{d}{dx}( 1 + cot x )\bigg)}{(1+cot x)^2}$

$\sf\dashrightarrow\dfrac{dy}{dx}= \dfrac{ 1 + cot x - x ( - cosec^2x ) }{( 1 + cot x)^2}$

$\sf\dashrightarrow\underset{\blue{\sf Required\ Derivative }}{\underbrace{\boxed{\pink{\sf \dfrac{dy}{dx}= \dfrac{x \ cosec^2x + cot\ x + 1 }{ ( 1 + cot x)^2} }}}}$

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