6. Differentiate the functions with respect to x in

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

Answers (1)

G Gautam harsolia

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 $10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5$