How do you differentiate #f(x)=(1+x^4)^(2/3)#?

1 Answer
VNVDVI
Feb 20, 2018

f'(x)= #(8/3x^3)(1+x^4)^(-1/3)#

Explanation:

When dealing with the chain rule, it's sometimes best to treat our composition of functions as "layered," where each layer must be differentiated individually and all differentiated layers must finally be multiplied. Here, our outermost layer would be the polynomial raised to the #2/3# power, and our innermost layer would be the polynomial #(1+x^4)#.

Differentiating the outer layer with the power rule, where the inner layer #(1+x^4)# has been temporarily denoted by "u" to avoid writing an incorrect answer, we would get:

#d/dx(u)^(2/3)=2/3(u)^(-1/3)#

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

Differentiating the inner layer with the power rule, we would get:

#d/dx(1+x^4)=4x^3#

Multiply our differentiated layers together and simplify:

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

f'(x)= #(8/3x^3)(1+x^4)^(-1/3)#

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