Question

sin[cos{tan(cotx)}] find dy/dx

Answer 1

what are u asking....

Answer 2

Answer: $\dfrac{dy}{dx}= cos(cos(tan(cotx))) (sin(tan(cotx)))(sec^2(cotx))(cosec^2x)$

Step-by-step explanation:

Let y = sin(cos(tan(cotx)))

Differentiating both side with respect to x we get

$\dfrac{dy}{dx} =\dfrac{d}{dy} sin(cos(tan(cotx)))\\\\\\\Rightarrow \dfrac{dy}{dx} = cos(cos(tan(cotx)))\dfrac{d}{dx} cos(tan(cotx))\\\\\\\Rightarrow \dfrac{dy}{dx} = cos(cos(tan(cotx))) (-sin(tan(cotx))) \dfrac{d}{dx}tan(cotx)\\\\\\\Rightarrow \dfrac{d}{dx} =cos(cos(tan(cotx))) (-sin(tan(cotx)))(sec^2(cotx)) \dfrac{d}{dx} cotx\\\\\\\Rightarrow \dfrac{dy}{dx}= cos(cos(tan(cotx))) (-sin(tan(cotx)))sec^2(cotx)( -cosec^2x)$

Hence, the $\dfrac{dy}{dx}= cos(cos(tan(cotx))) (sin(tan(cotx)))(sec^2(cotx))(cosec^2x)$