Why does dp/dy equal p' when substituting p for dy/dx?

[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 can't see the meaning of derivative by y

??

 

[foxjwill]

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

$$\begin{align*}<br /> \frac{dp}{dy} &=\frac{dp\bigl(y(x)\bigr)}{dy(x)} \\<br /> &= \frac{dp\bigl(y(x)\bigr)}{dy(x)} \cdot \frac{dx}{dx} \\<br /> &= \frac{dp\bigl(y(x)\bigr)}{dx} \biggl/ \frac{dy(x)}{dx}\\<br /> &= \frac{dp}{dx}\biggl/\frac{dy}{dx}\\<br /> &= \frac{y''(x)}{y'(x)}.<br /> \end{align*}$$​