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Find the deriative of r^(e^r) with respect to r(r>0)
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\[r^(e^r)\]
Use the chain rule with u=x and v=e^x\[\frac{d}{dx}\left( x ^{e^x} \right)=\frac{du^v}{du}\frac{du}{dx}+\frac{du^v}{dv}\frac{dv}{dx}\]where\[\frac{du^v}{du}=u ^{-1+v};\ \ \frac{du^v}{dv}=u^vln(u):\]\[=e^xx ^{e^x-1}\left( \frac{d}{dx}(x) \right)+x ^{e^x}\ln(x)\left( \frac{d}{dx}(e^x) \right)\]The derivative of x is 1...\[=x ^{e^x}\ln(x)\left( \frac{d}{dx}(e^x) \right)+1e^xx ^{e^x-1}\]The derivative of e^x is e^x...\[=e^xx ^{e^x-1+e^xx ^{e^xln(x)}}=e^xx ^{e^x-1}(xln(x)+1)\]
ohh... thanks :D
You are welcome... that was a haul!
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