How do you differentiate $y = x^{ln x}$ $y=x^lnx$ ?

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Answer 1 · 2016-04-18T01:46:34

$y'=2x^(ln(x)-1)ln(x)$

Using logarithmic and implicit differentiation, take the natural logarithm of both sides.

$ln(y)=ln(x^ln(x))$

Simplify the right hand side using the rule: $ln(a^b)=b*ln(a)$

$ln(y)=ln(x)*ln(x)$

$ln(y)=(ln(x))^2$

Take the derivative of both sides. Recall that the chain rule is in effect on both sides.

$(y')/y=2ln(x)overbrace(d/dx(ln(x)))^(=1//x)$

$(y')/y=(2ln(x))/x$

Now, multiply both sides by $y=x^ln(x)$ to solve for $y'$ .

$y'=(2x^ln(x)ln(x))/x$

Note that we can simplify this be writing $x^ln(x)/x$ as $x^ln(x)/x^1=x^(ln(x)-1)$ . Thus,

$y'=2x^(ln(x)-1)ln(x)$

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