# How do you calculate the derivative of the function f(x)=cos(x^3+x^2+1)?

Mar 6, 2015

Use the chain rule.

d/dx(f(g(x))=f'(g(x))g'(x)

In this problem, the outermost function ($f$) is the cosine function.

The derivative of the cosine function is the opposite of the sign function.

So we take $f '$ (the opposite of the sine) evaluated at the inside function ($g \left(x\right)$ which is ${x}^{3} + {x}^{2} + 1$) TIMES the derivative of the inside function.

The derivative is $- \sin \left({x}^{3} + {x}^{2} + 1\right)$ times $\left(3 {x}^{2} + 2 x\right)$ (the derivative of the inside function).

$f ' \left(x\right) = \left(- \sin \left({x}^{3} + {x}^{2} + 1\right)\right) \cdot \left(3 {x}^{2} + 2 x\right) = - \left(3 {x}^{2} + 2 x\right) \sin \left({x}^{3} + {x}^{2} + 1\right)$.