# Derivatives wrt functions

Gold Member

## Main Question or Discussion Point

Is it possible to take a derivative with respect to a function, rather than just a variable? I'll give a simple example of how I imagine such a thing would work to try to explain...

$$\frac{d}{d(sin(x))}\left(sin^2(x)\right) = 2 sin(x)$$

Can you take a derivative this way?

Also, can you write an equivalent integral as such?

$$\int 2 sin(x) d(sin(x)) = sin^2(x) + C$$

alxm
Sounds like you're looking for http://en.wikipedia.org/wiki/Functional_derivative" [Broken].

Last edited by a moderator:
To differentiate f(x) w.r.t g(x), just do the following:

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

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

$$=\frac{d f(x)}{d x}\left(\frac{d g(x)}{d x}\right)^{-1}$$
Char. Limit, for the integration, what you have written is the standard way I learnt to do integration. The general form is:

$$\int g(x) f'(x)d x=\int g(x)d f(x)$$

For example, by using the property of differential, $$d x^2/2 = x d x$$

$$\int e^{x^2}x \text{dx}$$

$$=\frac{1}{2}\int e^{x^2}\text{dx}^2$$

$$=\frac{1}{2}e^{x^2}+C$$

This method of doing integration is much better than using substitution in many situations. Since if the integral is complicated, you don't have to do substitution repeatedly.

Sounds like you're looking for http://en.wikipedia.org/wiki/Functional_derivative" [Broken].
no, it is just parametric ordinary differentiation.

$$w = f(u) = u^{2}, u(x) = \sin{(x)} \Rightarrow \frac{dw}{du} = f'(u) = 2 u = 2 \sin{(x)}$$

Functional derivatives are derivatives of functionals with respect to functions. $\sin^{2}{(x)}$ is not a functional.

Last edited by a moderator:
alxm
Functional derivatives are derivatives of functionals with respect to functions. $\sin^{2}{(x)}$ is not a functional.
Ah yes, silly me.. Guess I've been reading too many DFT papers lately.

Pengwuino
Gold Member
You'll even see things such as integrals over volumes, $$d^3x$$, done in terms of $$d(cos(\theta))$$ instead of just $$d\theta$$. So instead of $$r^2sin(\theta)drd\theta d\phi$$, you can change your limits of integration to be $$r^2drd(cos(\theta))d\phi$$. My first encounter, and I bet a lot of physics majors first encounter, is integrating Legendre Polynomials that are dependent on $$cos(\theta)$$ so it is only natural.

It's basically the method of substitution "on the go".

HallsofIvy
$$\frac{df}{dg}= \frac{df}{dx}\frac{dx}{df}= \frac{\frac{df}{ex}}{\frac{dg}{dx}}$$.