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.

  • 1
    $\begingroup$ Why does $U$ need to contain $c$? Is that supposed to say $U\ni f(c)$? $\endgroup$ – Sandejo Sep 12 '20 at 3:10
  • $\begingroup$ Is that g(f(x)) or g(x)×f(x) ???? $\endgroup$ – Soumyadwip Chanda Sep 12 '20 at 3:16
  • $\begingroup$ @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. $\endgroup$ – ΑΘΩ Sep 12 '20 at 4:44

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


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