Q2 Differentiate w.r.t. x the function in Exercises 1 to 11.

      \sin ^3 x + \cos ^6 x

Answers (1)

Given function is
f(x)= \sin ^3 x + \cos ^6 x
Now, differentiation w.r.t. x is
f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}
                                                                                =3\sin^2x.\cos x+6\cos^5x.(-\sin x)
                                                                                =3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)

Therefore, differentiation w.r.t. x is   3\sin x\cos x(\sin x- 2\cos ^4x)
 

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