# How do you differentiate f(x)=(1/x^3)sinx using the product rule?

May 25, 2018

$- 3 {x}^{-} 4 \sin x + {x}^{-} 3 \cos x$

#### Explanation:

Let's start by rewriting $f \left(x\right)$ as ${x}^{- 3} \sin x$. This allows us to differentiate it more easily.

We understand that if we have two functions, $g \left(x\right)$ and $h \left(x\right)$ being multiplied, the derivative is equal to

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

$f \left(x\right)$ is essentially composed of the following:

• $\textcolor{b l u e}{g \left(x\right) = {x}^{-} 3}$
• $\textcolor{red}{h \left(x\right) = \sin x}$

We can use the power rule and our knowledge of trig derivatives to figure out that

• $\textcolor{p u r p \le}{g ' \left(x\right) = - 3 {x}^{-} 4}$
• $\textcolor{\lim e}{h ' \left(x\right) = \cos x}$

Now, we just plug this business into our expression for the product rule. We get

$\textcolor{p u r p \le}{- 3 {x}^{-} 4} \textcolor{red}{\left(\sin x\right)} + \textcolor{b l u e}{{x}^{-} 3} \textcolor{\lim e}{\left(\cos x\right)}$

And I can write this with neutral colors to get

$\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} - 3 {x}^{-} 4 \sin x + {x}^{-} 3 \cos x \textcolor{w h i t e}{\frac{2}{2}} |}}$

Hope this helps!