How do you differentiate f(x)=(ln(x-(e^(tan(x^2)))))^(3/2) using the chain rule?
1 Answer
Explanation:
This is a mathematical bloodbath.
First Issue: the
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:
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
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:
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))
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)