# When is the derivative of $f(g(x))$ equal to $g(f'(x))$?

By the chain rule, we know that the derivative of $$f(g(x))$$ is $$f'(g(x))g'(x)$$.

Question: When is the derivative of $$f(g(x))$$ equal to $$g(f'(x))$$?

Trivial solutions include the following:

• Let $$f$$ be any differentiable function and $$g$$ be the identity function ($$g(x)=x$$) or the zero function ($$g(x)=0$$).
• Let $$f$$ be any constant function and $$g$$ be any differentiable function that fixes zero.

On the other hand, $$f(x)=x^2$$ and $$g(x)=\sin x$$ form a nontrivial solution (by the double angle formula for sine).

If $$g(x)=x+a$$, where $$a$$ is a constant, then $$f'(x+a)=f'(x)+a$$, so $$f(x)=\frac{1}{2}x^2+bx+c$$ would work for any two constants $$b$$ and $$c$$.

Finally, if $$f(x)=mx+b$$, where $$m$$ and $$b$$ are constants, then $$mg'(x)=g(m)$$, so $$g'(x)=\frac{g(m)}{m}$$ and $$g(x)=\frac{g(m)}{m}x+c$$ for some constant $$c$$. But this must in particular be true for $$x=m$$, so $$c$$ must be zero and $$g(x)=nx$$ would then work for any constant $$n$$.

But I do not know any other solutions. Can anyone help find one?

Note that $$f$$ may be replaced by any other function with the same derivative (i.e. differing from $$f$$ by a constant) without changing the validity of the equation, so we may assume without lost of generality that $$f$$ fixes zero (assuming, of course, that zero is in the domain of $$f$$).

• Do you know the more general definition of derivative as linear transformation? – Matematleta Sep 3 at 1:18