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Thursday, December 4, 2014

A one variable application of the multivariable chain rule

In the last post the theme was using the one variable chain rule in place of the multivariable version. I wanted to relay a pretty cool trick that goes in the other direction. Suppose you want to differentiate x^x. You remember that \frac{d}{dx} x^k=kx^{k-1} and \frac{d}{dx}b^x=\ln b b^x. Now think of x^x=u^v where u=v=x. The chain rule now says
\begin{align*} \frac{d}{dx} u^v&=(\ln u) u^v u_x+v u^{v-1}v_x\\ &=(\ln x) x^x+ x\, x^{x-1}\\ &=(\ln x+1)x^x \end{align*}
And we get the right answer as if by magic. In general, you can take a derivative by isolating every occurrence of the variable x, then differentiating while thinking of all other occurrences as constant, and then adding up the results.

2 comments:

  1. Is the final result correct? I guess it should read (\log x + 1)x^x

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