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$1)$ How do I denote derivative of $ax^2+b$ in terms of $ax^2$?

$(ax^2+b)'(ax^2)$ can easily be confused with $ax^2\cdot(ax^2+b)'$.

$2)$ How do I denote the derivative of $ax^2+b$ in terms of $ax^2$ at point $c$?

$3)$ How do I denote the derivative of $ax^2+b$ in terms of $x$ at point $c$?

$3)$ How do I denote the derivative of $f(g(x))$ in terms of $g(x)$ at point $a$?

I want to denote all of this without using $\text{d}$.

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  • $\begingroup$ Are you talking about applications of the chain rule? I'm not sure what you mean by "in terms of". $\endgroup$
    – GFauxPas
    Mar 24, 2015 at 1:22
  • $\begingroup$ @GFauxPas Yes. If $ax^2+b$ is a function of $ax^2$, then I'm searching how to denote its derivative. $\endgroup$
    – user89167
    Mar 24, 2015 at 1:24
  • $\begingroup$ By "in term of $ax^2$" you mean "with respect to the variable $ax^2$"? $\endgroup$ Mar 24, 2015 at 1:26
  • $\begingroup$ @DouglasFinamore If $f(x)=ax^2+b$ and $g(x)=ax^2$, then I'm searching how to denote $f'(g(x))$ without using $f$ and $g$. $\endgroup$
    – user89167
    Mar 24, 2015 at 1:27
  • $\begingroup$ Perhaps you mean "with respect to $x^2$"? Saying "with respect to $ax^2$" is unusual, if $a$ is a constant. $\endgroup$
    – GFauxPas
    Mar 24, 2015 at 1:28

1 Answer 1

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1) How do I denote derivative of $ax^2 +b$ in terms of $ax^2$ ?

$(ax^2 +b) ′ (ax^2 )$ can easily be confused with $ax^2 \cdot (ax^2 +b) ′$ .

Ah. Where as the prime notation on a function symbol is taken as being with respect to the function's argument, the prime notation over an expression is taken as being with respect to the independent variable of the discussion (most usually that is either $x$ or $t$).

That is, if $f(x)=ax+b$, then $\;f'(ax^2) = \frac{\mathrm d f(ax^2)}{\mathrm d (ax^2)} = \frac{\mathrm d (ax^2+b)}{\mathrm d (ax^2)} \\[2ex] f(ax^2)' = (ax^2+b)' = \frac{\mathrm d (ax^2+b)}{\mathrm d x}$

So, you want to use the prime notation on an expression to indicate you are deriving with respect to another expression rather than the implicit variable itself.

$$[u\mapsto u+b]' (ax^2) = \left.\frac{\mathrm d u+b}{\mathrm d u}\right\vert_{u:=ax^2} = \frac{\mathrm d(ax^2+b)}{\mathrm d (ax^2)}$$


You could establish in your forward that you were using a subscripted dash notation.

$$(ax^2+b)'_{(ax^2)} = \frac{\mathrm d (ax^2+b)}{\mathrm d (ax^2)}$$


Or simply rely on the chain rule. $\frac{\mathrm d (ax^2+b)}{\mathrm d x}\frac{\mathrm d x}{\mathrm d (ax^2)} = \frac{(ax^2+b)'}{(ax^2)'}$

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  • $\begingroup$ Your idea seems to be to simply replace $f$ with $[x\to x+b]$, which is sort of cheating. $\endgroup$
    – user89167
    Mar 24, 2015 at 11:00

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