Use the inverse function theorem to find the derivative of g(x)=x3\displaystyle {g{{\left({x}\right)}}}={\sqrt[{{3}}]{{{x}}}}.

g′(x)=1f′(g(x))\displaystyle {g}'{\left({x}\right)}=\frac{{1}}{{{f}'{\left({g{{\left({x}\right)}}}\right)}}}, where f(x)=g−1(x)\displaystyle {f{{\left({x}\right)}}}={{g}^{{-{1}}}{\left({x}\right)}}

Step 1 f(x)=g−1(x)=\displaystyle {f{{\left({x}\right)}}}={{g}^{{-{1}}}{\left({x}\right)}}=  
Step 2 f′(x)=\displaystyle {f}'{\left({x}\right)}=  
Step 3 f′(g(x))=\displaystyle {f}'{\left({g{{\left({x}\right)}}}\right)}=  
Step 4 1f′(g(x))=\displaystyle \frac{{1}}{{{f}'{\left({g{{\left({x}\right)}}}\right)}}}=  

Now, find g′(x)\displaystyle {g}'{\left({x}\right)} by differntiating g(x)\displaystyle {g{{\left({x}\right)}}} using the power rule, since x3=x13\displaystyle {\sqrt[{{3}}]{{{x}}}}={x}^{{\frac{{1}}{{3}}}}.
g′(x)=\displaystyle {g}'{\left({x}\right)}=  

License