# If f(x)= 9sin(sinx^2), then what is f^'(x)?

Jul 29, 2015

${f}^{'} \left(x\right) = 18 x \cdot \cos {x}^{2} \cdot \cos \left(\sin {x}^{2}\right)$

#### Explanation:

Start by breaking down your function - this will give you an idea about what you need to use to differentiate it

$f \left(x\right) = y = 9 \cdot \sin \left(\sin \left({x}^{2}\right)\right)$

From the looks of it, you're going to need to use the chain rule twice to differentiate $\sin \left(\sin \left({x}^{2}\right)\right)$.

The chain rule tells you that you can differentiate a function that is actually the composition of two other functions like this

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(y\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)}$, where

$y$ is a function that depends on variable $u$, which in turn depends on variable $x$.

Another important thing to remember is that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

So, you have $\sin \left(\sin {x}^{2}\right)$ as part of your original function. First, take $\sin \left({x}^{2}\right)$ to be $u$ so that you have

$y = 9 \cdot \sin u$

This can be differentiated like this

$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{du}} \left(9 \cdot \sin u\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(y\right) = 9 \cdot \frac{d}{\mathrm{du}} \left(\sin u\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

Now you need to use the chain rule for a second time, since you have

$u = \sin {x}^{2}$

This time, you can write $u = \sin {u}_{1}$, where ${u}_{1} = {x}^{2}$. This will get you

$\frac{d}{\mathrm{dx}} u = \frac{d}{{\mathrm{du}}_{1}} \cdot \sin {u}_{1} \cdot \frac{d}{\mathrm{dx}} {u}_{1}$

$\frac{d}{\mathrm{dx}} \left(u\right) = \cos {u}_{1} \cdot 2 x$, which is equivalent to

$\frac{d}{\mathrm{dx}} \left(u\right) = \cos {x}^{2} \cdot 2 x$

$\frac{d}{\mathrm{dx}} \left(y\right) = 9 \cdot \cos \left(\sin {x}^{2}\right) \cdot 2 x \cdot \cos {x}^{2}$
$\frac{d}{\mathrm{dx}} \left(y\right) = \textcolor{g r e e n}{18 x \cdot \cos {x}^{2} \cdot \cos \left(\sin {x}^{2}\right)}$