Derivative latex help

1. Aug 8, 2004

devious_

I posted the same thread twice. Oops. :tongue2:

Last edited: Aug 8, 2004
2. Aug 8, 2004

devious_

Derivative

$$x = e^u$$, where u is a function of x.

Using the chain rule:
$$\frac{dy}{du} = e^u\frac{dy}{dx}$$

Using the product rule:
$$\frac{d^2y}{du^2} = \frac{d}{du}(e^u\frac{dy}{dx}) = e^u\frac{dy}{dx}+e^u\frac{d^2y}{dx^2}\cdot\frac{dx}{du}$$

Why is it $$\frac{dx}{du}$$?

Last edited: Aug 8, 2004
3. Aug 8, 2004

arildno

What's this "y"-thingy?
It doesn't appear in your first line

4. Aug 8, 2004

Hurkyl

Staff Emeritus
(latex hint: you can use [ itex ] tags for formulas that go in a paragraph)

(Did you mean $y = e^u$?)

Anyways, the chain rule says that:

$$\frac{dp}{dq} = \frac{dp}{dr} \frac{dr}{dq}$$

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

So, throw it into the chain rule and see what you get.

5. Aug 8, 2004

devious_

Bleh, I'm new to latex so I accidentally pressed the post thread button instead of the preview post one.

Anyway, let me elaborate.

$$x = e^u$$, where u is a function of x.

Using the chain rule:
$$\frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du} = e^u \frac{dy}{dx}$$

Now, using the product rule:
$$\frac{d^2y}{du^2} = \frac{d}{du}(\frac{dy}{du}) = \frac{d}{du}(e^u \frac{dy}{dx}) = e^u \frac{dy}{dx} + e^u \frac{d^2}{dx^2} \cdot \frac{dx}{du}$$

My question is:
Shouldn't $\frac{dx}{du}$ be $\frac{dy}{du}$?

Last edited: Aug 8, 2004
6. Aug 8, 2004

devious_

Nevermind. I see where I went wrong.