Derivative

[devious_]

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

[devious_]

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.