Question

differentiate w. r.t x: sin^2 (x^5)​

Answer 1

Answer: 10x⁴cosx⁵.sinx⁵

To find:

$\sf \dfrac{d}{dx}sin^2(x^5)$

Formula used:

$\boxed{\sf Chain\:rule: \dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}}$

$\boxed{\sf Power\:rule:(x^n)=nx^{n-1}}$

Explanation:

$\sf Let$

$\sf y=sin^2(x^5)$

$\sf u=x^5$

$\sf \therefore y=sin^2 u$

$\sf Hence,\:\dfrac{du}{dx}=5x^{5-1}=5x^4$

$\sf and\:\dfrac{dy}{du}=sin^2u$

$\sf Again\:using\:chain\:rule\:to\:differentiate\:sin^2u,we\:get:$

$\sf \therefore\dfrac{d}{du}sin^2u=2\:sinu.cosu$

$\sf Substituting\:the\:value\:of\:u,we\:get:$

$\sf \dfrac{dy}{dx}=2\:sinx^5.cosx^5.5x^4$

     $\sf =\boxed{\sf10x^4cosx^5.sinx^5}$