How do you use the chain rule to differentiate #y=1/(2x^3+7)#?

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Answer 1 · 2016-10-22T12:43:54

the derivative is #dy/dx=(-6x^2)/((2x^2+7)^2)#

Let #u(x)=2x^3+7#

then #y=(u(x))^(-1)#

so the derivative is #dy/dx=-1(u(x))^(-2)*u'(x)#

we use the following #(x^n)'=nx^(n-1)#

and #u'(x)=(2x^3+7)'=6x^2#

So the final result is #(-1/(2x^3+7)^2)*6x^2#