# What is the derivative of g(x)=x^3 cos x?

Mar 25, 2018

$g ' \left(x\right) = 3 {x}^{2} \cos x - {x}^{3} \sin x$

#### Explanation:

Since $g \left(x\right)$ is the product of two terms, we can use the Product Rule to find the derivative.

We essentially have $g \left(x\right) = f \left(x\right) \cdot h \left(x\right)$, where

$\textcolor{p u r p \le}{f \left(x\right) = {x}^{3}}$ and

$\textcolor{g r e e n}{h \left(x\right) = \cos x}$

Thus, the Product Rule states that the derivative is equal to:

$f ' \left(x\right) h \left(x\right) + f \left(x\right) h ' \left(x\right)$

To differentiate $f \left(x\right)$, we can use the Power Rule, where the exponent becomes the coefficient, and we decrement the power. Thus,

$\textcolor{b l u e}{f ' \left(x\right) = 3 {x}^{2}}$

And from our knowledge of derivatives of trig functions

$\textcolor{red}{h ' \left(x\right) = - \sin x}$

We can now plug these values into the product rule expression to get

$\textcolor{b l u e}{3 {x}^{2}} \textcolor{g r e e n}{\left(\cos x\right)} + \textcolor{p u r p \le}{{x}^{3}} \textcolor{red}{\left(- \sin x\right)}$

We can rewrite this as

$3 {x}^{2} \left(\cos x\right) - {x}^{3} \left(\sin x\right)$

Thus, $g ' \left(x\right) = 3 {x}^{2} \cos x - {x}^{3} \sin x$

If the Power or Product Rules seem foreign to you, I encourage you to Google them or go to Khan Academy to understand them more.

Hope this helps!