# Derivative with respect to something else

I have a problem that I'm stuck on. In my calculus class we've only learned to take the derivative with respect to x, y, or time. The question is;

y = x^2 + x. The derivate of y with respect to 1/(1-x) is .. ? If someone could help me get started, specifically how to get started it'd be appreciated. Thanks.

## Answers and Replies

Fredrik
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This is how a physicist would do it:

$$\frac{d}{d(\frac{1}{1-x})}=\frac{dx}{d(\frac{1}{1-x})}\frac{d}{dx}=\frac{1}{\frac{d(\frac{1}{1-x})}{dx}}\frac{d}{dx}=\frac{1}{\frac{1}{(1-x)^2}}\frac{d}{dx}=(1-x)^2\frac{d}{dx}$$

The interpretation of this is that to take the derivative with respect to 1/(1-x) is the same as taking the derivative with respect to x and then multiplying the result with (1-x)^2.

This is another way to do it:

$$f(x)=x^2+x$$

Let g be the function that satisfies

$$f(x)=g\big(\frac{1}{1-x}\big)=g(h(x))$$

The definition of h(x) is of course just h(x)=1/(1-x).

What we're looking for is

$$g'(h(x))$$.

Note that $$f=g\circ h$$.

$$f'(x)=(g\circ h)'(x)=g'(h(x))h'(x)$$

$$g'(h(x))=\frac{f'(x)}{h'(x)}=(1-x)^2 f'(x)$$

The second approach has the advantage that it's much easier to explain every step.

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Fredrik
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I decided to give some thought to what I was actually doing when I did the first calculation, and I came to the conclusion that this is what the first calculation actually says:

$$g'(h(x))=(f\circ h^{-1})'(h(x))$$

$$=f'(h^{-1}(h(x)))h^{-1}'(h(x))=h^{-1}'(h(x))f'(x)$$

$$=\frac{1}{h'(h^{-1}(h(x)))}f'(x)=\frac{1}{h'(x)}f'(x)$$

$$=\frac{1}{\frac{1}{(1-x)^2}}f'(x)$$

$$=(1-x)^2 f'(x)$$

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Gib Z
Homework Helper
Im not the best at this, but i would let what you want with respect to, equal u, and then find what y equals in terms of u.

Im not the best at this, but i would let what you want with respect to, equal u, and then find what y equals in terms of u.

This is what I would do to, and just to clarify, this is usually referred to as "making a subsitution". Making the substitiution $u = \frac{1}{1-x}$ and then finding $\frac{dy}{du}$ will solve the problem. It also makes it quite a lot easier to write y in terms of u since we can rearrange the expression defining u:

$$\frac{1}{1-x} = u$$
$${1-x} = \frac{1}{u}$$
$${-1+x} = \frac{-1}{u}$$
$$x = \frac{-1}{u} + 1$$ or,
$$x = 1 - \frac{1}{u}$$

Which can then be substituted back into the original

$$y = x^2 + x$$
$$y = \left(1 - \frac{1}{u}\right)^2 + \left(1 - \frac{1}{u}\right)$$
$$y = 2 - \frac{3}{u} + \frac{1}{u^2}$$

Which, barring any mathematical errors on my part, should allow you to find the derivative $\frac{dy}{du}$ as required. Once found just substitute back the original expression we had for u: $u = \frac{1}{1-x}$

This is of course identical to using h(x) in place of u and g(x) from the above posts would be the new function for y in terms of u which is given above. The only reason I chose to use the u is that it is tidier to write and makes it "more obvious" what you should do at each stage, since it takes the form of a variable, rather than the less familiar form of a function. Personally I get confused when I see something like $f'(g(x))$ : is it the derivative of f with respect to x, or with respect to g(x)?

Fredrik
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Personally I get confused when I see something like $f'(g(x))$ : is it the derivative of f with respect to x, or with respect to g(x)?
It doesn't really make sense to say that the derivative of a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is with respect to a variable with a specified name. There's only one variable, and it doesn't matter if we call it x, y, t or even g(x).

It only makes sense to specify what the derivative is with respect to when we haven't already specified what function we're taking the derivative of. For example, it wouldn't make sense to talk about "the derivative of f(g(x))", because f(g(x)) isn't a function, it's a number. But if we say that the derivative is with respect to x, we make it clear that the function we're taking the derivative of is the map $x\mapsto f(g(x))$ (also known as $f\circ g$). If we say that the derivative is with respect to g(x), we're specifying that the function we're taking the derivative of is the map $y\mapsto f(y)$ (also known as f).

f'(g(x)) is the derivative of f, evaluated at g(x). Another way of saying that is that f'(g(x)) is the derivative of f(g(x)) with respect to g(x), evaluated at g(x).

But I guess I'm nitpicking here. Your post is very good. A lot of people probably think your explanation is much easier to understand than mine, and my advice to the thread starter is to try to understand both methods.

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