Question

Find the derivative of cos(log(cotx)) w.r.t 'x'.​

Answer 1

Topic :-

Differentiation

To Differentiate :-

y = cos(log(cotx)) with respect to 'x'.

Solution :-

$y=\cos(\ln(\cot x))$

$\dfrac{dy}{dx}=\dfrac{d(\cos(\ln(\cot x)))}{dx}$

$\dfrac{dy}{dx}=-\sin(\ln(\cot x))\cdot \dfrac{d(\ln(\cot x))}{dx}$

$\left ( \because \dfrac{d(\cos t)}{dx}=-\sin t\cdot\dfrac{dt}{dx} \right)$

$\dfrac{dy}{dx}=-\sin(\ln(\cot x))\cdot \dfrac{1}{\cot x}\cdot \dfrac{d(\cot x)}{dx}$

$\left ( \because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{dy}{dx}=-\sin(\ln(\cot x))\cdot \dfrac{1}{\cot x}\cdot (-\csc^2x)$

$\left ( \because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{dy}{dx}=\dfrac{\sin(\ln(\cot x))\cdot \csc^2x}{\cot x}$

Answer :-

Required derivative is :-

$\boxed {\dfrac{dy}{dx}=\dfrac{\sin(\ln(\cot x))\cdot \csc^2x}{\cot x}}$

Answer 2

Answer:

sin[log(cotx)].cosec²x

Step-by-step explanation:

let u=log(cotx). v=cotx

u=logv. y=sinu

dy/dx=d/du cosu×d/dv logv×d/dx cotx

= - sinu × 1/v × - cosec²

= - ( sinu × -1/v × cosec²x)

= - (sin log (cotx)× -1/cotx × cosec²x)

= {sin[log(cotx)]× cosec²x}\cotx