Can I denote derivative of $f(g(x))$ in terms of $g(x)$ by $f'(g(x))_{g(x)}$?

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How to denote derivative of $f(g(x))$ in terms of $g(x)$ in prime (with $'$ without $\text{d}$) notation?

Is it conventional to denote derivative of, say, $\sin(\cos x)$ in terms of $\cos x$ in the following way: $$\sin'(\cos x)_{\cos x}?$$

This notation is useful for using chain rule: $$f'(g(x))=f'(g(x))_{g(x)}g'(x)$$

Edit: Please explain your downvote. If it is nonsensical to denote it this way, then at least say so in the comments.

Edit2: I've posted a follow-up question on differentiation notation.

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Write \begin{align*} (f\circ g)(x)=f\big(g(x)\big) \end{align*} and then \begin{align*} (f\circ g)'(x)=f'\big(g(x)\big)g'(x) \end{align*} or \begin{align*} (f\circ g)'(x)=f'(y)g'(x)\qquad\text{where}\qquad y=g(x). \end{align*} The expression \begin{align*} f'\big(g(x)\big) \end{align*} means the derivative of $f$ evaluated at $g(x)$, not the derivative of the function composition $f\big(g(x)\big)$ evaluated at $x$.