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champ
 5 years ago
what is the derivative of
x^x2
champ
 5 years ago
what is the derivative of x^x2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the second x raised to the 2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is a bit tricky and requires a little bit of thinking. Firstly, to clarify.. you wand \(\frac{d}{dx}x^{x^2}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i think that's what he means, because had it been multilication he/she would have just put the two in front..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you going to solve this polpak? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Crazy! How's this look? <a href=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d72914b5ebe30d944f1a0ea34/20110422_1313.png "><img class="embeddedObject" src=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d72914b5ebe30d944f1a0ea34/20110422_1313.png " width="596" height="653" border="0" /></a>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure. 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
 5 years ago
Best ResponseYou've already chosen the best response.0i dont uderstand how you got to ln

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I took the ln(u) to get the x^2 out of the exponent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If \(u = x^{x^2}\) then \(ln(u) = x^2ln(x)\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you use a formula on the rest of it am kinda lost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No formula. Just the chain rule and the product rule.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The right side of the equation on line 3 is the chain rule, the left is the product rule.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you made a small error x^2/x=x not 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I had the 1 there because I was gonna factor the x from both terms, then forgot to do it ;p

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0the equation still holds if polak takes the natural log of both sides then you use implicit differentiation to solve for y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so when trying to deferentiate x^2*lnx you use the product rule
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