Question

solve it !!!????!! ??​

Answer 1

Answer:

$\bigstar\;\boxed{\sf \dfrac{dy}{dx}=\left[\dfrac{-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)}{\Bigg( x^{2}\Bigg)}\right]}$

Explanation:

$\rule{300}{1.5}$

Given equation,

$\longrightarrow\sf y=\cot\Bigg(\pi-\dfrac{1}{x}\Bigg)$

Differentiating it,

$\longrightarrow\sf \dfrac{dy}{dx}=\dfrac{d}{dx}\Bigg[\cot\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]$

Applying chain rule we get,

$\longrightarrow\sf \dfrac{dy}{dx}=\dfrac{d}{dx}\Bigg[\cot\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\dfrac{d}{dx}\Bigg\lgroup \pi-\dfrac{1}{x}\Bigg\rgroup$

Derivative of cot is - cosec².

$\longrightarrow\sf \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\dfrac{d}{dx}\Bigg\lgroup \pi-\dfrac{1}{x}\Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\Bigg\lgroup\dfrac{d(\pi)}{dx}-\dfrac{d\;(x^{-1})}{dx}\Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\Bigg\lgroup0-\Bigg(-1\times x^{-1-1}\Bigg)\Bigg\rgroup$

π is a constant, differentiating it will give zero.

$\sf\longrightarrow \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\Bigg\lgroup0-\Bigg(-x^{-2}\Bigg)\Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\Bigg\lgroup x^{-2}\Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\Bigg[-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)\Bigg]\times\Bigg\lgroup \dfrac{1}{x^2}\Bigg\rgroup\\\\\\\\\longrightarrow\sf \dfrac{dy}{dx}=\left[\dfrac{-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)}{x^{2}}\right]$

$\longrightarrow \underline{\boxed{\red{\sf \dfrac{dy}{dx}=\left[\dfrac{-\csc^2\Bigg(\pi-\dfrac{1}{x}\Bigg)}{\Bigg( x^{2}\Bigg)}\right]}}}$

Hence Solved!

$\rule{300}{1.5}$