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Simplified to a very basic problem, is there a standard procedure for these types of differentiations?

$$\displaylines{ y = {x^{{e^x}}} \cr {{d} \over {dx}}\left( y \right) = {{d} \over {dx}}\left( {{x^{{e^x}}}} \right) \cr = {x^{{e^x}}}\left( {{e^x}\ln \left( x \right) + {{{e^x}} \over x}} \right) \cr} $$

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What you have is correct. The standard procedures you probably already know: power rule, product rule, quotient rule, chain rule, derivatives of the basic functions. Also, rewriting $x^{f(x)}=e^{f(x)\ln x}$ crops up sometimes. What else do you want? –  anon Mar 20 '13 at 6:07

2 Answers 2

up vote 1 down vote accepted

Here is an approach.

Take logarithm on both sides, $$\log y=\log x^{e^x}=e^x\log x$$ Differentiate both sides, $$\frac{y'}{y}=e^x\log x+\frac{e^x}{x}$$ Multiply both sides by $y$ and substitute the original equation $$\frac{dy}{dx}=y'=(e^x\log x+\frac{e^x}{x})y=(e^x\log x+\frac{e^x}{x})x^{e^x}$$

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I like to leave something for OP to think about. –  Gerry Myerson Mar 20 '13 at 6:07
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@GerryMyerson I think this trick is taught in all high schools. Did you come up with this idea on your own? –  NECing Mar 20 '13 at 6:08
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The idea of logarithmic differentiation? No, I didn't come up with it on my own. But I might have worked it out if someone had told me, take logs & differentiate --- and I would have had a feeling of accomplishment if I had been able to work it out from that. –  Gerry Myerson Mar 20 '13 at 6:11
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@GerryMyerson I didn't post the actual problem I am working on for precisely that reason. Thanks. –  Patrick Mar 20 '13 at 6:12

Hint: take logarithms on both sides, then differentiate, recalling $(\log y)'=y'/y$.

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