If f(x)= cot2 x f(x)=cot2x and g(x) = e^(1 - 4x ) g(x)=e14x, how do you differentiate f(g(x)) f(g(x)) using the chain rule?

1 Answer
Oct 13, 2016

(8e^(1-4x))/sin^2(2e^(1-4x))8e14xsin2(2e14x) or 8e^(1-4x)csc^2(2e(1-4x))8e14xcsc2(2e(14x))

Explanation:

f(g(x))=cot2e^(1-4x)f(g(x))=cot2e14x

Let g(x)=ug(x)=u

f'(u)=d/(du)cot2u=d/(du)(cos2u)/(sin2u)=(-2sin(2u)sin(2u)-2cos(2u)cos(2u))/sin^2(2u)
=(-2sin^2(2u)-2cos^2(2u))/sin^2(2u)
=-2/sin^2(2u)

g'(x)=-4e^(1-4x)

Using chain rule: f'(g(x))=f'(u)*g'(x)

=-2/sin^2(2u)*-4e^(1-4x)

=-2/sin^2(2e^(1-4x))*-4e^(1-4x)

=(8e^(1-4x))/sin^2(2e^(1-4x)) or 8e^(1-4x)csc^2(2e(1-4x))