How do you differentiate # y= 4/(x^3+1)^(2/3)# using the chain rule?

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Answer 1 · 2017-07-31T08:18:51

The derivative is #=-(8x^2)/(x^3+1)^(5/2)#

We need

#(x^n)'=nx^(n-1)#

Our function is

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

Therefore,

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

#=-(8x^2)/(x^3+1)^(5/2)#