If $g'(x)$ is constant, what is the fifth derivative of $f(g(x))$?

Question

I just need some clarification about how to solve and approach this problem, we did it in a Calc 1474 class but I didn't understand how to approach and solve it.

Answer 1

You can use the chain rule $h(x)=f(g(x))$ implies that $h'(x)=f'(g(x))g'(x)$.

Then, chain rule and product rule: $h''(x)=f''(g(x))(g'(x))^2+f'(g(x))g''(x)$. Since $g'$ is constant then $g''(x)=0$ for all $x$. Therefore $h''(x)=f''(g(x))(g'(x))^2$.

You can continue in the same way (notice that whenever you try to compute the second term in the product rule you get a $g''$ which is zero and the term dissapears) to get that:

\begin{align}h'''(x)&=f'''(g(x))(g'(x))^3\\ h^{(iv)}(x)&=f^{(iv)}(g(x))(g'(x))^4\\ h^{(v)}(x)&=f^{(v)}(g(x))(g'(x))^5\end{align}

Answer 2

Let $g'(x) = c$. By chain rule we have $$f(g(x))'=f'(g(x))g'(x) = cf'(g(x)).$$ Now, since $c$ is constant, if we take derivative again, we get $$f(g(x))''=(cf'(g(x)))' = c(f'(g(x)))'=c^2f''(g(x)).$$ At this point, one might guess that $$f(g(x))^{(n)}=c^nf^{(n)}(g(x))$$ is true. Indeed, we can use induction to prove it: $$f(g(x))^{(n+1)}=(c^nf^{(n)}(g(x)))'=c^n(f^{(n)}(g(x)))' = c^{n+1}f^{(n+1)}(g(x)).$$

Now, let $n=5$ to finish the exercise.

Answer 3

The first derivative is $f'(g(x))(g'(x))$, the second is $f''(g(x))(g'(x))^2$, here you apply the formula of the derivative of a product taking in account that $(g')'=0$ recursively, the fith is $f^{(5)}(g(x))(g'(x)))^5$.