6. Differentiate the functions with respect to x in 

\cos x^3 . \sin ^ 2 ( x ^5 )

Answers (1)

Given function is
f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )
Differentitation w.r.t. x is
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)
g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)
Lets take u = x^3 \ and \ v = x^5
Our functions become,
\cos x^3 = \cos u   and    \sin^2(x^5) = \sin^2v
Now,
g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx}                        ( By chain rule)
\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)
\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2
g^{'}(x) = -\sin x^3.3x^2                            -(i)
Similarly,
h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}
\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5 \ \ \ (\because v = x^5)
                                                                                                                      
\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4
h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5                             -(ii)
Put (i) and (ii) in
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) = -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5\cos x^5
Therefore, the answer is 10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5 

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