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

1 Answer
Nov 6, 2017

Answer:

#f'(x)=-(3x^2sec^2 (x^3-2))/e^(tan(x^3-2))#

Explanation:

We have #f(x)=e^(tan(2-x^3))#. We can see that #f# is a composition of three different functions, so in order to find #f'#, we need to differentiate each of these three functions as if their arguments were a single variable and then multiply their derivatives.

#d/dxe^u=e^u#

#d/dxtanv=sec^2v#

#d/dx(2-x^3)=-3x^2#

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

We have the derivative, but it doesn't look particularly nice, so we can rewrite it to look neater.

#sec(-theta)=sectheta rArrsec(2-x^3)=sec(x^3-2)#

#tan(-theta)=-tanthetarArrtan(2-x^3)=-tan(x^3-2)#

#e^(-u)=1/e^urArre^tan(2-x^3)=e^(-tan(x^3-2))=1/e^tan(x^3-2)#

#therefore f'(x)=-(3x^2sec^2 (x^3-2))/e^(tan(x^3-2))#