# Thread: Differentiating with respect to a variable squared?

1. ## Differentiating with respect to a variable squared?

I have the following:

$\frac{\delta W}{\delta I_4}$

Where $I_4 = \alpha^2$

My question is, does $\frac{\delta W}{\delta \alpha^2} = \frac{\delta W}{\delta \alpha}$?

2. Originally Posted by Bwts
I have the following:

$\frac{\delta W}{\delta I_4}$

Where $I_4 = \alpha^2$

My question is, does $\frac{\delta W}{\delta \alpha^2} = \frac{\delta W}{\delta \alpha}$?

ps I have no idea why my latex code isnt working on your site?
Try keeping up to date with what happens by reading the posts here: http://www.mathhelpforum.com/math-help/f26/ ! (I have applied the temporary fix to your latex).

3. Originally Posted by Bwts
I have the following:

$\frac{\delta W}{\delta I_4}$

Where $I_4 = \alpha^2$

My question is, does $\frac{\delta W}{\delta \alpha^2} = \frac{\delta W}{\delta \alpha}$?

[snip]
Use the chain rule.

4. It does not give you the same answer, if you differentiate with respect to a variable and with respect to that variable squared.
For example, consider the function f(x)=x^2. If you take the derivative of this function with respect to x, you of course get 2x. If you take the derivative with respect to x^2 (which is df/dx^2), then you get the answer 1. Why? In df/dx^2, you treat x^2 as a single variable, so you might as well call it something less confusing, say v=x^2. Then the problem of finding df/dx^2, when f(x)=x^2, translates to finding df/dv, when f(v) = v, which is 1.

Similarly, g(x)=x^4 is 4x^3 upon differentiation with respect to x, but with respect to x^2, it becomes 2x^2 (since x^4 = (x^2)^2, and (x^2)^2 differentiated with respect to x^2 is 2x^2).

5. I'm thinking that maybe you have an easier time relating to Fantastic's suggestion.

By the chain rule, we have that

df/dx = df/dx^2 * dx^2/dx,

but since dx^2/dx = x^2, we have that df/dx^2 =(df/dx)/(2x). So to find the derivative with respect to x^2, you find the derivative of f with respect to x, and then divides this answer by 2x.

6. Thanks that makes sense now.