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I'm feeling a bit brain-dead, perhaps it has just been a long day. But here is the problem, I think it's a simple one:

I have a function $y=f(x)$, of which I then take the derivative $\frac{dy}{dx}$. Finally I take the derivative of the function with respect to itself:

$$ \frac{d}{dy}\left(\frac{dy}{dx}\right)$$

How can I simplify/refactor this funky derivative? Ideally to express the derivative in terms of $y$ and derivatives thereof.

Elliot Gorokhovsky
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ibell
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3 Answers3

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There's no need to get all fancy... We want to solve for $\frac{dy'}{dy}$ and by the chain rule we have

$$\frac{dy'}{dx} = \frac{dy}{dx}\frac{dy'}{dy}$$ so, solving for our quantity, we get $$\frac{dy'}{dy} = \frac{\frac{dy'}{dx}}{\frac{dy}{dx}}=\frac{y''}{y'}$$

Elliot Gorokhovsky
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The chain rule might suggest an answer something like $\dfrac{\frac{d^2y}{dx^2}}{\frac{dy}{dx}}$

For example with $y=x^n$ you might get $\dfrac{n(n-1)x^{n-2}}{nx^{n-1}}=\dfrac{n-1}{x}$. To check this, at least for positive $x$, $\frac{dy}{dx}=nx^{n-1}=ny^{(n-1)/n}$ so $\frac{d}{dy}\left(\frac{dy}{dx}\right) = (n-1)y^{-1/n}=\frac{n-1}{x}$

On the other hand for $y=ce^{kx}$ you might just get the constant $k$. To check this, $\frac{dy}{dx}=kce^{kx}=ky$ so $\frac{d}{dy}\left(\frac{dy}{dx}\right) = k$ as expected

Curiously for $y=\tan(x/2)$ you get the same $\tan(x/2)$

Henry
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$dy/dx=f'(x)$ is a function of $x$, just as $f$ is. In order to differentiate the function $f'(x)$ with respect to $y$, a variable related to $x$, it must somehow be expressed as a function of $y$.

An analogy which may be helpful: $x^2-4$ is a function of $x$. Its derivative with respect to $x$ is $2x$, i.e., the function which takes $x$ to $2x$. Suppose we wrote $y=x^2$. Then $\dfrac{d}{dy}(x^2-4)=$ $\dfrac{d}{dy}(y-4)=1$, a constant.

Introduce the notation $f^{-1}$ for the inverse function of $f$ over some suitable interval $I$ in which $f$ is increasing or decreasing (which is required for the inverse to exist). Then $x=f^{-1}(y)$ for $y\in f(I)$, and so $f'(x)=f'(f^{-1}(y))$ for $y\in f(I)$. Having thus expressed $f'$ as a function of $y$, we can calculate its derivative with respect to $y$. Differentiate via the chain rule (Baby Rudin 5.5) and get $\dfrac{d}{dy}f'(x)=\dfrac{d}{dy}f'(f^{-1}(y))=f''(f^{-1}(y))\dfrac{d}{dy}f^{-1}(y)$. Now use Baby Rudin exercise 5.2 to get an expression for the second factor in this last expression.

ForgotALot
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    +1: I feel like this is the only answer which carefully explains the situation at hand. I like to think of this using manifolds: $f':\mathbb{R} \to \mathbb{R}$ is the function to be differentiated, and $y = f(x)$ is coordinate system we are using on the domain. As any chart, $f$ should be injective (at least in a small enough neighborhood). – Alex Provost Jul 26 '16 at 03:26