Question

find the derivative with respect to x
y= cosec (2sin x)
use chain rule
​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = cosec(2sinx)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} cosec(2sinx)$

We know

$\boxed{\tt{ \dfrac{d}{dx}f[g(x)] = f'[g(x)]\dfrac{d}{dx}g(x)}}$

and

$\boxed{\tt{ \dfrac{d}{dx}cosecx \: = \: - \: cosecx \: cotx}}$

So, on substituting the values, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - cosec(2sinx)cot(2sinx)\dfrac{d}{dx}(2sinx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = -2 cosec(2sinx)cot(2sinx)\dfrac{d}{dx}(sinx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = -2 cosec(2sinx) \: cot(2sinx) \: cosx$

Hence,

$\rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = -2 cosec(2sinx) \: cot(2sinx) \: cosx}}$

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MORE TO KNOW

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$