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

1 Answer
Jan 25, 2016

Answer:

#f'(x) = (3 [ln( 1 - e^(-cos (x^2)) )]^(-5/2) e^(-cos(x^2)) * sin x^2 * x )/ ( 1 - e^(-cos (x^2)) )#

Explanation:

1) Simplifying:

Let me simplify the expression a little bit before differentiating:

You can use the power rule

#a^(-m) = 1 / a^m #

to transform your expression as follows:

#f(x) = 1 / [ln(1 - e^(-cos (x^2)) ) ]^(3/2) = [ln(1 - e^(-cos (x^2)) ) ]^(-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(x) = [ln( 1 - e^(-cos (x^2)) )]^(-3/2) = [color(orange)(ln( 1 - e^(-cos (x^2)) ))]^(-3/2) = color(orange)(z)^(-3/2)#

where

#z = ln( 1 - e^(-cos (x^2)) ) = ln(color(blue)( 1 - e^(-cos (x^2))) ) = ln(color(blue)(u))#

where

# u = 1 - e^(-cos (x^2)) = 1 - e^(color(violet)(-cos (x^2))) = 1 - e^color(violet)(v)#

where

# v = -cos (x^2) = - cos(color(red)(x^2)) = - cos(color(red)(w))#

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'(x) = [z^(-3/2)]' * z' * u' * v' * w' #

3) Computing derivatives:

Let's compute the five derivatives:

# [z^(-3/2)]' = -3/2 z^(-5/2) = -3/2 [ln( 1 - e^(-cos (x^2)) )]^(-5/2)#

# [z]' = [ln(u)]' = 1/u = 1/ ( 1 - e^(-cos (x^2)) ) #

# [u]' = [1 - e^v]' = - e^v = - e^(-cos(x^2))#

# [v]' = [-cos w]' = sin w = sin x^2 #

# [x^2]' = 2x#

4) Applying the chain rule:

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

#f'(x) = [z^(-3/2)]' * z' * u' * v' * w' #

#= -3/2 [ln( 1 - e^(-cos (x^2)) )]^(-5/2) * 1/ ( 1 - e^(-cos (x^2)) ) * (- e^(-cos(x^2))) * sin x^2 * 2x#

#= 3/cancel(2) * [ln( 1 - e^(-cos (x^2)) )]^(-5/2) * 1/ ( 1 - e^(-cos (x^2)) ) * e^(-cos(x^2)) * sin x^2 * cancel(2) x#

#= (3 [ln( 1 - e^(-cos (x^2)) )]^(-5/2) e^(-cos(x^2)) * sin x^2 * x )/ ( 1 - e^(-cos (x^2)) ) #