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

1 Answer
Jan 3, 2016

Answer:

#f'(x)=((6xe^tan(x^2)sec^2(x^2)-3)sqrt(ln(x-e^tan(x^2))))/(2e^tan(x^2)-2x)#

Explanation:

This is a mathematical bloodbath.

First Issue: the #3/2# power. Deal with through chain rule and power rule: #d/dx(u^(3/2))=3/2u^(1/2)=3/2sqrtu#.

#f'(x)=3/2sqrt(ln(x-e^tan(x^2)))*d/dx(ln(x-e^tan(x^2)))#

Second Issue: the natural logarithm. Use the chain rule: #d/dx(ln(u))=1/u*u'#.

#f'(x)=3/2sqrt(ln(x-e^tan(x^2)))*1/(x-e^tan(x^2))*d/dx(x-e^tan(x^2))#

A little simplification, first...

#f'(x)=(3sqrt(ln(x-e^tan(x^2))))/(2(x-e^tan(x^2)))*d/dx(x-e^tan(x^2))#

Third Issue: the #e# with the tangent power. Use the chain rule: #d/dx(e^u)=e^u*u'#.

#f'(x)=(3sqrt(ln(x-e^tan(x^2))))/(2(x-e^tan(x^2))) * (1-e^tan(x^2)d/dx(tan(x^2)))#

Fourth Issue: the tangent. Use chain rule again: #d/dx(tan(u))=u'sec^2(u)#.

#f'(x)=(3sqrt(ln(x-e^tan(x^2))))/(2(x-e^tan(x^2))) * (1-e^tan(x^2)sec^2(x^2)d/dx(x^2))#

#d/dx(x^2)# should be very easy to find!

#f'(x)=(3sqrt(ln(x-e^tan(x^2))))/(2(x-e^tan(x^2))) * (1-2xsec^2(x^2)e^tan(x^2))#

Some simplification can be done to reach...

#f'(x)=((6xe^tan(x^2)sec^2(x^2)-3)sqrt(ln(x-e^tan(x^2))))/(2e^tan(x^2)-2x)#