# Homework Help: Why p''=pp' ?

1. Jul 27, 2009

### electron2

why p''=pp' ??

i have a function
y(x)

i want to substitute by variable p.

p=y'(x)=dy/dx

every derivative is by dx because its the smallest variable
so why

why dp/dy=p'

i cant see the meaning of derivative by y

??

2. Jul 27, 2009

### foxjwill

Re: why p''=pp' ??

First of all, writing dp/dy=p' implies that you're differentiating with respect to y. Otherwise, you are right, dp/dy is not equal to dp/dx in this case.

The expression $$\frac{dp}{dy}$$ is really just a short-hand for the chain rule. The best thing to do here is not to think of p as a variable but instead as a function p(y(x))=y'(x). What it's asking is "What happens to the value of p(y(x)) as the value of y(x) changes?" That is, we want to find
$$\frac{{\rm change\ in\ }p\bigl(y(x)\bigr)}{{\rm change\ in\ }y(x)},$$​
or, in "semi-short-hand" notation,
$$\frac{dp\bigl(y(x)\bigr)}{dy(x)},$$​
and in standard short-hand notation,
$$\frac{dp}{dy}.$$​
Now, since both p(y(x)) and y(x) are in terms of x, we can re-write these (albeit with a slight abuse of notation) as
\begin{align*} \frac{dp}{dy} &=\frac{dp\bigl(y(x)\bigr)}{dy(x)} \\ &= \frac{dp\bigl(y(x)\bigr)}{dy(x)} \cdot \frac{dx}{dx} \\ &= \frac{dp\bigl(y(x)\bigr)}{dx} \biggl/ \frac{dy(x)}{dx}\\ &= \frac{dp}{dx}\biggl/\frac{dy}{dx}\\ &= \frac{y''(x)}{y'(x)}. \end{align*}​