LaTeX derivative help

[devious_]

I posted the same thread twice. Oops.

 

[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 $$\frac{dx}{du}$$?

 

[arildno]

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

 

[Hurkyl]

(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:

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

In your calculation, you had to compute:

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

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

 

[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 $\frac{dy}{du}$?

 

[devious_]

Nevermind. I see where I went wrong.