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))#
