If #f(x)= 1/x # and #g(x) = x^3 #, how do you differentiate #f'(g(x)) # using the chain rule?

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Answer 1 · 2017-11-01T10:32:03

#-(3)/x^4#

#f(x) = 1/x# and #g(x) = x^3#

For #f(g(x))# we use #1/(g(x))= 1/x^3#

Derivative of #f(g(x))# using chain rule:

Let #u= x#

Then.

#dy/dx = dy/(du) * (du)/dx#

#dy/dx = dy/(du)(u^-3) * (du)/dx(u)= -3u^-4*1=(-3)/x^4#