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,

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

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

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