Can I use chain rule for 2nd derivative?

Question

If I do have $(x^{4}+10x^{2}+1)^{98}$ and I have to find 2nd derivative of it, can I just simply do that like:

$(f\circ g)''(x) = f''(g(x))*g''(x)$ ?

Answer 1

No. We have $$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$$ so $$\frac{d^2}{dx^2}f(g(x))=f''(g(x))\cdot g'(x) \cdot g'(x)+ f'(g(x))\cdot g''(x)\neq f''(g(x))\cdot g''(x)$$

So in your example, $$f(x)=x^{98}\implies f'(x)=98x^{97}\implies f''(x)=98\cdot97x^{96}$$ and $$g(x)=x^4+10x^2+1\implies g'(x)=4x^3+20x\implies g''(x)=12x^2+20$$ so $$\frac{d^2}{dx^2}(x^4+10x^2+1)^{98}=\cdots?$$

Answer 2

No, since $$(f\circ g)'=(f'\circ g)g'$$ then differentiating again gives $$(f\circ g)''=(f''\circ g)g'^2+(f'\circ g)g''.$$

See Faa di Bruno's formula for higher derivatives.

Answer 3

The first derivative of $$x\mapsto f(g(x))$$ is (by the chain rule) $$x\mapsto g'(x)f'(g(x)).$$ So to obtain the derivative of this (i.e., the second derivative of $f\circ g$), you need the product rule and the chain rule $$ x\mapsto g''(x)f'(g(x)+g'(x)\cdot g'(x)f''(g(x))$$