Conservation of Critical Points Under Strictly Monotone Function Composition

Suppose we have a function $$f: \mathbb{R} \rightarrow \mathbb{R}$$, which is differentiable on some interval $$I$$. Let's say $$\exists c \in I: f'(c) = 0$$.

If we have a strictly monotone function $$g:U \rightarrow \mathbb{R}$$, where $$U$$ is some open subset of $$\mathbb{R}$$ containing $$f(c)$$, why does $$(g$$$$f)'(c) = 0?$$

I have tried reasoning this using the limit definition of the derivative, and applying definitions for monotonicity, but can't seem to reach a conclusion.

• Why does $U$ need to contain $c$? Is that supposed to say $U\ni f(c)$? – Sandejo Sep 12 '20 at 3:10
• Is that g(f(x)) or g(x)×f(x) ???? – Soumyadwip Chanda Sep 12 '20 at 3:16
• @Raiyan Chowdhury Beware that in order to be able to legitimately speak about the composition $g \circ f$ you must require the codomain $\mathbb{R}$ of $f$ to coincide with the definition domain $U$ of $g$, so $U$ can't be just ''some open subset''. Either that or you are thinking about the composition between $g$ and a suitable restriction of $f$, whose codomain would be $U$. At any rate, without precise ammendments, the notation $g \circ f$ is illicit. – ΑΘΩ Sep 12 '20 at 4:44

Derivative rule for compound functions: $$(g\circ f)'(c)=g'(f(c))\cdot f'(c)$$.