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

Jan 25, 2016

$f ' \left(x\right) = \frac{3 {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{5}{2}} {e}^{- \cos \left({x}^{2}\right)} \cdot \sin {x}^{2} \cdot x}{1 - {e}^{- \cos \left({x}^{2}\right)}}$

#### Explanation:

1) Simplifying:

Let me simplify the expression a little bit before differentiating:

You can use the power rule

${a}^{- m} = \frac{1}{a} ^ m$

to transform your expression as follows:

$f \left(x\right) = \frac{1}{\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)} ^ \left(\frac{3}{2}\right) = {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{3}{2}}$

Now, let's apply the chain rule:

2) Defining the chain:

Try to break down the function into a chain of composed functions:

$f \left(x\right) = {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{3}{2}} = {\left[\textcolor{\mathmr{and} a n \ge}{\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)}\right]}^{- \frac{3}{2}} = {\textcolor{\mathmr{and} a n \ge}{z}}^{- \frac{3}{2}}$

where

$z = \ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right) = \ln \left(\textcolor{b l u e}{1 - {e}^{- \cos \left({x}^{2}\right)}}\right) = \ln \left(\textcolor{b l u e}{u}\right)$

where

$u = 1 - {e}^{- \cos \left({x}^{2}\right)} = 1 - {e}^{\textcolor{v i o \le t}{- \cos \left({x}^{2}\right)}} = 1 - {e}^{\textcolor{v i o \le t}{v}}$

where

$v = - \cos \left({x}^{2}\right) = - \cos \left(\textcolor{red}{{x}^{2}}\right) = - \cos \left(\textcolor{red}{w}\right)$

where, finally,

$w = {x}^{2}$

So, after you've split up your function in a chain of composed functions, your derivative can be computed with the following product:

$f ' \left(x\right) = \left[{z}^{- \frac{3}{2}}\right] ' \cdot z ' \cdot u ' \cdot v ' \cdot w '$

3) Computing derivatives:

Let's compute the five derivatives:

$\left[{z}^{- \frac{3}{2}}\right] ' = - \frac{3}{2} {z}^{- \frac{5}{2}} = - \frac{3}{2} {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{5}{2}}$

$\left[z\right] ' = \left[\ln \left(u\right)\right] ' = \frac{1}{u} = \frac{1}{1 - {e}^{- \cos \left({x}^{2}\right)}}$

$\left[u\right] ' = \left[1 - {e}^{v}\right] ' = - {e}^{v} = - {e}^{- \cos \left({x}^{2}\right)}$

$\left[v\right] ' = \left[- \cos w\right] ' = \sin w = \sin {x}^{2}$

$\left[{x}^{2}\right] ' = 2 x$

4) Applying the chain rule:

Now, we have all the parts and can apply the rule:

$f ' \left(x\right) = \left[{z}^{- \frac{3}{2}}\right] ' \cdot z ' \cdot u ' \cdot v ' \cdot w '$

$= - \frac{3}{2} {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{5}{2}} \cdot \frac{1}{1 - {e}^{- \cos \left({x}^{2}\right)}} \cdot \left(- {e}^{- \cos \left({x}^{2}\right)}\right) \cdot \sin {x}^{2} \cdot 2 x$

$= \frac{3}{\cancel{2}} \cdot {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{5}{2}} \cdot \frac{1}{1 - {e}^{- \cos \left({x}^{2}\right)}} \cdot {e}^{- \cos \left({x}^{2}\right)} \cdot \sin {x}^{2} \cdot \cancel{2} x$

$= \frac{3 {\left[\ln \left(1 - {e}^{- \cos \left({x}^{2}\right)}\right)\right]}^{- \frac{5}{2}} {e}^{- \cos \left({x}^{2}\right)} \cdot \sin {x}^{2} \cdot x}{1 - {e}^{- \cos \left({x}^{2}\right)}}$