champ
  • champ
what is the derivative of x^x2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
is the second x raised to the 2?
anonymous
  • anonymous
i think its x(2x)
anonymous
  • anonymous
hmm, perhaps x^2x..

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

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