Let $g'(x)$ be constant. Find the fifth derivative of $f\circ g(x)$.

Question

I have a question with some of my calculus homework. Here's the question:
Let $g'(x)$ be constant. Find the fifth derivative of $f\circ g(x)$.
First, is $f\circ g(x)$ just $fg(x)$? And how do I go about starting this question? If $g'(x)$ is a constant, would that mean $g(x)$ equals a number?
Thanks in advance!

Answer 1

First, integrate $g'\left(x\right)=c_1$ to get $g\left(x\right)=c_1x+c_2$, where $c_1$ and $c_2$ are arbitrary constants. The composition $f\circ g\left(x\right)$ is then $f\left(c_1x+c_2\right)$. Let $h\left(x\right)=f\circ g\left(x\right)$. Differentiation of this is just repeated application of the chain rule: $$h'\left(x\right)=c_1f'\left(c_1x+c_2\right)$$ $$h''\left(x\right)=c_1^2f''\left(c_1x+c_2\right)$$ $$\vdots$$ $$h^{\left(5\right)}\left(x\right)=c_1^5f^{\left(5\right)}\left(c_1x+c_2\right)$$

Answer 2

The derivative of $f\circ g(x) = f(g(x))$ is $$f'(g(x))\cdot g'(x)$$ Can you take it from here?

Hint: You can take another derivative and not worry about taking the derivative of $g'(x)$ because it is a constant! See Chain Rule.

Answer 3

We have $$(f\circ g)'(x)=f'(g(x))\cdot g'(x)$$ by the Chain Rule. Let $g'(x)=c$ for some constant $c.$ Thus, we have $$(f\circ g)'(x)=f'(g(x))\cdot g'(x)=c\cdot f'(g(x)).$$ Claim that $$(f\circ g)^{(n)}=c^n\cdot f^{(n)}(g(x))$$ for all integers $n>0.$ Clearly, as we have shown above, this holds for $n=1.$ Now for the inductive step. Assume that $$(f\circ g)^{(n)}=c^n\cdot f^{(n)}(g(x)).$$ Then, we have \begin{align*} ((f\circ g)^{(n)})'&=(c^n\cdot f^{(n)}(g(x)))'\\ &=(c^n)'\cdot f^{(n)}(g(x))+c^n\cdot (f^{(n)}(g(x)))' \end{align*} by the Product Rule. Since $(c^n)'=0,$ we have \begin{align*} (c^n)'\cdot f^{(n)}(g(x))+c^n\cdot (f^{(n)}(g(x)))'&=0(f^{(n)}(g(x)))+c^n\cdot f^{(n+1)}(g(x))\cdot g'(n)\\ &= c^{n+1}\cdot f^{(n+1)}(g(x)) \end{align*} by the Chain Rule and that $g'$ is constant, which is want we want. Thus the inductive step is done. Putting in $n=5,$ the fifth derivative of $(f\circ g)$ is $c^5\cdot f^{(5)}(g(x))=\boxed{(g'(x))^5\cdot f^{(5)}(g(x))}.$ $\blacksquare$