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I am presented with the following task:

"Assume that the function $f(x)$ has the derivative $f'(x) = \frac{1}{x}$ and that $f$ is one-to-one. If $y = f^{-1}(x)$, show that $\frac{dy}{dx} = 1$.

The solution given in the same textbook starts out with the following statement: $$x = f(y) \rightarrow f'(y)*\frac{dy}{dx} = 1$$

From this point, the proof is trivial, but I am confused by the notation given. Would $f'(y)*\frac{dy}{dx}$ translate to "the derivative of the function $f(y)$ with respect to $x$? If so, could you elaborate?

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That's correct. If $x = f(y)$, then differentiating both sides with respect to $x$ we have $$1 = \frac{d}{dx}(f(y)) = \frac{df}{dx}(y)\frac{dy}{dx} = f'(y)\frac{dy}{dx}$$ where the equality between the second and third terms comes from the chain rule.

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Classical notation has an allure but is wrought with potential for confusion.

Let $\phi(x) = f^{-1}(x)$. Then $f(\phi(x)) = x$. Then $D f(\phi(x)) D \phi(x) = 1$. (Note, so far this has nothing to do with the actual value of $f'$).

Returning to classical notation, with $y=\phi(x)$, then $f'(y) \frac{dy}{dx} = 1$.

To complete, we are given that $Df(x) = \frac{1}{x}$, hence we have $\frac{1}{\phi(x)} D\phi(x) = 1$, or $D \phi(x) = \phi(x)$, or more classically, $\phi' = \phi$. From this we see that $\phi(x) = \phi_0 e^x$, and since this gives $x = \log \frac{\phi(x)}{\phi_0}$, we see that $f(x) = \log \frac{x}{\phi_0}$.

My suggestion: Use an unambiguous notation for calculation, and revert to the classical notation for results and mnemonics. In the long run, the extra paper used will be more than compensated for by correct answers and clarity of understanding.

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The book uses the chain rule $\dfrac{d(f(y))}{dx} = \dfrac{df}{dy}*\dfrac{dy}{dx}$, where $\dfrac{df}{dy} = f'(y)$. So this is the result when you take the derivative of both sides. Yes, this is "the derivative of the function $f(y)$ with respect to x" using the chain rule.

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