How do you use the chain rule to differentiate #y=1/(x^4-1)#?

1 Answer
Steve M
Oct 18, 2016

# dy/dx= (-4x^3)/(x^4-1)^2 #

Explanation:

We have, # y = 1/(x^4-1) #

So, let # u=x^4-1 # then:
# (du)/dx = 4x^3 #

And then, # y=1/u = u^-1 => dy/(du)=(-1)u^-2#
# :. dy/(du)=-1/u^2 #

The chain rule tells us that; # dy/dx = dy/(du) (du)/dx #

Applying this gives us:
# dy/dx = -1/u^2 4x^3 #

# = -1/(x^4-1)^2 4x^3 #
# = (-4x^3)/(x^4-1)^2 #

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