# How do you differentiate f(x)=sin((x/(x^2-1))^(3/2)) using the chain rule?

Nov 26, 2015

$f ' \left(x\right) = \frac{3}{2} \cos \left({\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{3}{2}}\right) {\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{1}{2}} \cdot \frac{- {x}^{2} - 1}{{x}^{2} - 1} ^ 2$

#### Explanation:

$f \left(x\right) = \sin \left({\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{3}{2}}\right)$

Ok, so let's break your function down into a chain of functions:

$f \left(u\right) = \sin \left(u\right)$

$u \left(v\right) = {v}^{\frac{3}{2}}$

$v \left(x\right) = \frac{x}{{x}^{2} - 1}$

As next, you need to differentiate those three functions.

$f \left(u\right) = \sin \left(u\right) \textcolor{w h i t e}{\times x} \implies f ' \left(u\right) = \cos \left(u\right)$

$u \left(v\right) = {v}^{\frac{3}{2}} \textcolor{w h i t e}{\times \times \times} \implies u ' \left(v\right) = \frac{3}{2} {v}^{\frac{1}{2}}$

To differentiate $v \left(x\right)$, you can use the quotient rule:

if $f \left(x\right) = g \frac{x}{h \left(x\right)}$, the derivative is $f ' \left(x\right) = \frac{g ' \left(x\right) \cdot h \left(x\right) - h ' \left(x\right) \cdot g \left(x\right)}{{h}^{2} \left(x\right)}$.

So, in your case, the derivative is:

$v \left(x\right) = \frac{x}{{x}^{2} - 1} \textcolor{w h i t e}{x} \implies v ' \left(x\right) = \frac{1 \cdot \left({x}^{2} - 1\right) - 2 x \cdot x}{{x}^{2} - 1} ^ 2 = \frac{- {x}^{2} - 1}{{x}^{2} - 1} ^ 2$

Now, to compute the derivative of $f \left(x\right)$, you just need to multiply your three derivatives! :-)

$f ' \left(x\right) = f ' \left(u \left(v \left(x\right)\right)\right) \cdot u ' \left(v \left(x\right)\right) \cdot v ' \left(x\right)$

$\textcolor{w h i t e}{\times \times} = \cos \left({\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{3}{2}}\right) \cdot \frac{3}{2} {\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{1}{2}} \cdot \frac{- {x}^{2} - 1}{{x}^{2} - 1} ^ 2$

$\textcolor{w h i t e}{\times \times} = \frac{3}{2} \cos \left({\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{3}{2}}\right) {\left(\frac{x}{{x}^{2} - 1}\right)}^{\frac{1}{2}} \cdot \frac{- {x}^{2} - 1}{{x}^{2} - 1} ^ 2$