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I searched on the internet and found the derivation $$\frac{dy}{d(ax)}=\frac{1}{a}\frac{dy}{dx}\tag{1}$$ I am confused of why this is true.

I tried to do the derivation myself, but there was no progress. Here is what I did. $$y(x)=2x$$ $$\frac{dy}{dx}=2$$ So I say that $$y(2x)=4x$$ According to $(1)$ $$\frac{dy}{d(2x)}= \frac{1}{2}\frac{dy}{dx}=1$$ But how does this even make sense?

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4 Answers 4

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Generally, when you have something like

$$\frac{d}{d(g(x))} f(x)$$

I personally find it convenient to make a substitution of $u = g(x)$ and then rewrite $f$ as a function of $u$.

Take note of your error: in your case, you just took the bottom, and plugged in $g(x)$ into $f(x)$ as an input. That's not at all how this sort of situation works. You're essentially checking the rate of change of $f$ with respect to the variable $g(x)$. Typically we just let $g(x) = x$, but that is not always the case, but it can be reduced to that case by manipulations and substitution.

For $f(x) = 2x$, you are correct in that $f'(x) = 2$. However, in finding

$$\frac{d}{d(2x)} 2x$$

we let $u = 2x$. Then $f(x) = 2x \Leftrightarrow f(u) = u$, and thus

$$\frac{d}{d(2x)} 2x = \frac{d}{du} u = 1 = \frac{1}{2} \frac{d}{dx} 2x$$

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This is simple if you realize that the infinitesimal differential operator $\text{d} f(x) = f'(x) \, \text{d}x$ where $f'(x)$ is the derivative of $f$ with respect to $x$.

In your case $f(x) = ax$ and thus $\text{d} f(x) = a \,\text{d} x$.

Thus, obtain $$ \frac{\text{d}y}{\text{d}(ax)}=\frac{\text{d}y}{a\,\text{d}x}, $$ as desired.

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I know people helped you already, buy maybe this will make it more clear.

using the chain rule: $$\frac{dy}{d(ax)}=\frac{dy}{dx}\frac{dx}{d(ax)}=\frac{dy}{dx}\frac{d(\frac{1}{a}(ax))}{d(ax)}=\frac{dy}{dx}\frac{1}{a}=\frac{1}{a}\frac{dy}{dx}$$

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$$\frac{d}{dx}f(x)=f'(x)$$ To solve for $$\frac{d}{d(ax)}f(x)=f'(x)$$ Make a substitution of $u=ax$, we have $$\frac{d}{du}f\left(\frac{u}{a}\right)=\frac{1}{a}\left(f'\left(\frac{u}{a}\right)\right)=\frac{1}{a}f'(x)=\frac{1}{a}\frac{d}{dx}$$

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  • $\begingroup$ Your first two equations imply $\frac{d}{dx}f(x)=\frac{d}{d(ax)}f(x)$, which is probably not what you intended. Also, are you missing an $f(x)$ at the end of your last equation? $\endgroup$
    – nwsteg
    Aug 31, 2022 at 23:48

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