Question

find derivtives of function x³- cosx ​

Answer 1

ANSWER:

$Here, \\ \: y= x^3-\cos x$

$\therefore \frac{d}{dx} y=\frac{d}{dx}x^3 - \frac{d}{dx}\cos x$

$Now,\\ \frac{d}{dx} x^n=nx^{n - 1}$

$And, \frac{d}{dx} \cos x= -\sin x$

$\therefore \frac {d}{dx} y = 3x^2 -(-\sin x)$

$\color {gold} \therefore \frac{d}{dx} y= 3x^2 + \sin x$