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Let's suppose I've got a function $f(x)$ where I'd like to differentiate with respect to $t$, but $t$ depends on $x$: $t(x)$. Thus the whole linked derivative thing: $\dfrac{\mathrm{d}f(x)}{\mathrm{d}t(x)}$. Is this possible at all? Alternatively I had to find $t^{-1}(x)$: $x(t)$ and then calculate the derivative fairly easy: $\mathrm{d} f(x(t))/\mathrm{d}t$. But finding the inverse is not always possible. Is there another way? At all?

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    $\begingroup$ Example?......... $\endgroup$
    – Randall
    Commented Oct 23, 2021 at 15:16

2 Answers 2

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Maybe this is what you're looking for:

$$ \frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt} = \frac{df}{dx} \frac{1}{\frac{dt}{dx}}. $$

For example: Consider $f(x) = x^2$ and $t = e^x$.

  1. Direct calculation of $\frac{df}{dt}$: Note that $x = \ln{t}$, so $f = \ln^2t$. Then using chain rule:

$$ \frac{df}{dt} = 2\ln{t} (\frac{1}{t}) = 2x \frac{1}{e^x}.$$

  1. Using the formula above:

$$\frac{df}{dt} = \frac{df}{dx} \frac{1}{\frac{dt}{dx}} = 2x \frac{1}{e^x}. $$

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$$ \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} = \frac{df}{dx}\left(\frac{dt}{dx}\right)^{-1} = \frac{f'(x)}{t'(x)}. $$

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