Question

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

Answer 1

`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)`