Question

\[
\frac{d}{d(x^2)}x
\]The derivative of \(x\) with respect to \(x^2\).

I'm getting \(1/(2x)\) but wolfram is getting just \(1/x\). I wonder how they are getting that.

Answer 1

how are you getting that?

Answer 2

In the general case: \[
\frac{d}{d(x^2)}x^n
\]I would say this is \[
\frac{n}{2}x^{n-2}
\]But wolfram does not get that \(n/2\) coefficient.

Answer 3

@agent0smith

Answer 4

@nincompoop I used substitution.

Answer 5

I know, link the wolfram

Answer 6

http://www.wolframalpha.com/input/?i=d%2F%5Bd%28x%5E2%29%5D+x

Answer 7

Simplify the following:
(d x)/(d x^2)
Cancel common terms in the numerator and denominator of (d x)/(d x^2).
(d x)/(d x^2) = d/d×x/x^2 = x/x^2:
x/x^2
For all exponents, a^n/a^m = a^(n-m). Apply this to x/x^2.
Combine powers. x/x^2 = x^(1-2):
x^1-2
Evaluate 1-2.
1-2 = -1:
Answer:
x^-1

Answer 8

that d is not coming out as derivative operation

Answer 9

Here is one way I considered doing it: \[
\frac{d}{d(x^2)}x = \frac{d}{2xdx}x = \frac{1}{2x}\frac{d}{dx}x = \frac{1}{2x}
\]

Answer 10

Another way: \(u = x^2\) and \(x = u^{1/2}\): \[
\frac{d}{d(x^2)}x = \frac{d}{du}x = \frac{d}{du}u^{1/2} = \frac{1}{2u^{1/2}} = \frac{1}{2x}
\]

Answer 11

http://www.wolframalpha.com/input/?i=differentiate+x+respect+to+x%5E2

no evaluation

Answer 12

Suppose \[
\frac{d}{d(x^2)}x = y
\]Then \[
dx = yd(x^2)=y(2x)dx \implies 1 = y(2x)\implies y = \frac{1}{2x}
\]

Answer 13

Is wolfram pelletting the bed?

Answer 14

laughing out loud
I am trying to get the right command
http://www.wolframalpha.com/input/?i=dy%2Fdx%5E2+x+

Answer 15

What does wolfram think it is doing when it gets \(1/x\)...

Answer 16

or \(x^{n-2}\) for that matter.

Answer 17

Hmm... http://www.wolframalpha.com/input/?i=d%2F%5Bd%28x%5E2%29%5D+x%5E4and x^3 does the same, there's no constant out the front.