what is the derivative of x^x2

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what is the derivative of x^x2

Mathematics
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is the second x raised to the 2?
i think its x(2x)
hmm, perhaps x^2x..

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This is a bit tricky and requires a little bit of thinking. Firstly, to clarify.. you wand \(\frac{d}{dx}x^{x^2}\)
Err want.
Right?
yeah, i think that's what he means, because had it been multilication he/she would have just put the two in front..
Are you going to solve this polpak? :)
Crazy! How's this look?
Sure. The original poster seems to have gone away, but it's a fun exercise in anycase. Let \(u = x^{x^2} \implies (ln\ u) = x^2(ln\ x)\) \[\frac{d}{dx}(ln\ u) = \frac{d}{dx}x^2(ln\ x)\] \[\implies \frac{d}{du}(ln\ u) * \frac{du}{dx} = x^2*\frac{d}{dx}(ln\ x) + (ln\ x)*\frac{d}{dx}(x^2)\] \[\implies \frac{1}{u} * \frac{du}{dx} = \frac{x^2}{x} + 2x(ln\ x) \] \[\implies \frac{du}{dx} = u(1 + 2x(ln\ x))\] \[ = x^{x^2}(1 + 2x(ln\ x)) = x^{x^2} + 2x^{x^2+1}(ln\ x)\] \[\implies \frac{d}{dx}x^{x^2} = x^{x^2} + 2x^{x^2+1}(ln\ x)\]
i dont uderstand how you got to ln
I took the ln(u) to get the x^2 out of the exponent.
If \(u = x^{x^2}\) then \(ln(u) = x^2ln(x)\)
did you use a formula on the rest of it am kinda lost
No formula. Just the chain rule and the product rule.
1 Attachment
The right side of the equation on line 3 is the chain rule, the left is the product rule.
you made a small error x^2/x=x not 1
Ah, so I did.
I had the 1 there because I was gonna factor the x from both terms, then forgot to do it ;p
the equation still holds if polak takes the natural log of both sides then you use implicit differentiation to solve for y'
so when trying to deferentiate x^2*lnx you use the product rule
yes

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