What are the first and second derivatives of #f(x)=ln(1-x)^(3/2) #?

Question

What are the first and second derivatives of #f(x)=ln(1-x)^(3/2)   #?

Emily Ammerman · Accepted Answer

Assuming that what you meant was #ln^(3/2)(1-x)#, we can apply chain rule twice here.

Where #f(x)=u^(3/2)#, #u=ln(v)# and #v=1-x#. #(df(x))/(dx)=(3u)^(1/2)/2(1/v)(-1)# Substituting #u#: #(df(x))/(dx)=-(3(ln(v))^(1/2))/(2v)# Substituting #v#: #(df(x))/(dx)=color(green)((3sqrt(ln(1-x)))/(2(x-1)))# The second derivative falls into the same logic for chain rule in the numerator. However, we have a fraction, which will demand quotient rule. #(df(x)^2)/(d^2x)=(((3/2u^(-1/2))(1/v)(-1))(2(x-1))-3sqrt(ln(1-x))(2)}/(4(x-1)^2)# Substituting #u#: #(df(x)^2)/(d^2x)=(((3/(2sqrt(ln(v)))(1/v)(-1))(2(x-1))-6sqrt(ln(1-x))))/(4(x-1)^2)# Substituting #v#: #(df(x)^2)/(d^2x)=(((3/(2sqrt(ln(1-x)))(1/(1-x))(-1))(2(x-1))-6sqrt(ln(1-x))))/(4(x-1)^2)# #(df(x)^2)/(d^2x)=((cancel(-)6cancel((x-1)))/(2cancel((1-x))sqrt(ln(x-1)))-6sqrt(ln(1-x)))/(4(x-1)^2)# #(df(x)^2)/(d^2x)=((3-6ln(1-x))/(sqrt(ln(1-x))))/(4(x-1)^2)=color(green)((3-6ln(1-x))/(4(x-1)^2sqrt(ln(1-x))))#

Paisley Johnson · Answer

undefined First derivative:
$ f'(x) = -\frac{3}{2(1-x)} $

Second derivative:
$ f''(x) = \frac{3}{2(1-x)^2} $