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

Answer 1

Isaiah Allen

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

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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Answer 2

Christopher Agresta

To differentiate ( y = \frac{1}{x^4 - 1} ) using the chain rule, follow these steps:

Rewrite the function as ( y = (x^4 - 1)^{-1} ).
Identify the inner function, ( u = x^4 - 1 ).
Find ( \frac{du}{dx} ), the derivative of the inner function with respect to ( x ), which is ( \frac{du}{dx} = 4x^3 ).
Apply the chain rule: ( \frac{dy}{dx} = -1 \times (x^4 - 1)^{-2} \times \frac{du}{dx} ).
Substitute ( \frac{du}{dx} ) and simplify to get the final derivative.

So, the derivative of ( y = \frac{1}{x^4 - 1} ) with respect to ( x ) is: [ \frac{dy}{dx} = \frac{-4x^3}{(x^4 - 1)^2} ]

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Answer from HIX Tutor

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