What is the second derivative of cube root of (x^2-1)

Question

owlcoffee · Answer

$$f:f(x)=\sqrt[3]{(x^2-1}$$
You can make use of the exponential property $\sqrt[n]{A}=A ^{\frac{ 1 }{ n }}$ which will simplify the derivation of this function thereby performin the transformation $f:f(x)\rightarrow f':f'(x)$ :
$$f:f(x)=(x^2-1)^{\frac{ 1 }{ 3 }}$$
$$f':f'(x)=\frac{ 1 }{ 3 }(x^2-1)^{(\frac{ 1 }{ 3 }-1)}(2x)$$
Remember that $f(x)=\sqrt[3]{x^2-1}$ is a composite function, because we don't have a positive "x" alone inside the root, which means we have to use the derivation rule: $H(x)=g(f(x)) \iff H'(x)=g'(f(x)).f'(x)$.

owlcoffee · Answer

after you operate that, you derivate again to obtain the second derivative.

anonymous · Answer

Refer to the attachment.