Question

Differentiate sin^2x with respect to e^cos x

Answer 1

=-2cosx/e^cosx
Where u=sin^2x. V=e^cosx
Differentiate with respect to x.
du/dx=2sinx.cosx. dv/dx=sinx.e^cosx
Finally
du/dx/dv/dx=du/dv=-2cosx/e^cosx

Answer 2

we have to differentiate sin²x with respect to $e^{cosx}$.

step1 : differentiate, y = sin²x with respect to x,

dy/dx = d(sin²x)/dx

= 2sinx × d(sinx)/dx

= 2sinx × cosx

hence, differentiation of sin²x with respect to x is dy/dx = 2sinx.cosx

step2 : differentiate z = e^{cosx} with respect to x,

dz/dx = d(e^{cosx})/dx

= e^{cosx} × d(cosx)/dx

= e^{cosx} × (-sinx)

= -sinx . e^{cosx}

hence, differentiation of e^{cosx} with respect to x is dz/dx = -sinx . e^{cosx}

step3 : now dividing (dy/dx) by (dz/dx), we get differentiation of sin²x with respect to e^(cosx).

i.e., (dy/dx)/(dz/dx) = (2sinx.cosx)/(-sinx.e^{cosx})

or, dy/dz = -2cosx/e^{cosx}

hence, differentiation of sin²x with respect to e^(cosx) is -2cosx/e^{cosx}