How do you find the derivative of $y=x^2+x^x$ ?

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Answer 1 · 2015-07-17T22:30:58

We know the first half. Let's try the second half, using a more foolproof way than simply trying to remember the derivative of a similar function.

Let:
$y = x^x$

$log_x(y) = log_x(x^(x)) = x$

$(lny)/(lnx) = x$

$lny = xlnx$

$1/y ((dy)/(dx)) = x*1/x + lnx*1$
(Implicit Differentiation, Product Rule)

$1/y ((dy)/(dx)) = 1 + lnx$

$((dy)/(dx))_(y = x^x) = y[1 + lnx]$

$= x^x[1 + lnx]$

Thus you have overall:

$color(blue)(((dy)/(dx))_(y = x^2 + x^x) = 2x + x^x(1+lnx))$

How do you find the derivative of y=x^2+x^xy=x^2+x^x?