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

1 Answer
Jan 3, 2016

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)