How to determine a derivative if the derivative is dependent itself?

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?

• Example?......... Commented Oct 23, 2021 at 15:16

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:

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

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