Q 8 \cos ( a \cos x + b \sin x ), for some constant a and b. 

Answers (1)

Given function is
f(x)=\cos ( a \cos x + b \sin x )
Now, differentiation w.r.t x
f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}
                                     = -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}
                                    = -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)
                                    = (a\sin x-b\cos x)\sin(a\cos x+b\sin x).
Therefore, differentiation w.r.t x  (a\sin x-b\cos x)\sin(a\cos x+b\sin x)

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