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

Question

1568 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-13T19:15:20

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

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

Let #g(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))#