How do you find the derivative of $y = x^{ln x}$ $y = x^(ln x)$ ?

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Answer 1 · 2016-10-14T02:44:38

Take the natural logarithm of both sides:

$ln y = ln (x^{ln x})$ $lny = ln(x^lnx)$

$ln y = ln x (ln x)$ $lny = lnx(lnx)$

$\frac{1}{y} (\frac{d y}{d x}) = \frac{1}{x} \times ln x + \frac{1}{x} \times ln x$ $1/y(dy/dx) = 1/x xx lnx + 1/x xx lnx$

$\frac{1}{y} (\frac{d y}{d x}) = \frac{ln x}{x} + \frac{ln x}{x}$ $1/y(dy/dx) = lnx/x + lnx/x$

$\frac{1}{y} (\frac{d y}{d x}) = \frac{2 ln x}{x}$ $1/y(dy/dx) = (2lnx)/x$

$\frac{d y}{d x} = \frac{\frac{2 ln x}{x}}{\frac{1}{y}}$ $dy/dx= ((2lnx)/x)/(1/y)$

$\frac{d y}{d x} = y \times \frac{2 ln x}{x}$ $dy/dx= y xx (2lnx)/x$

$\frac{d y}{d x} = x^{ln x} \times \frac{2 ln x}{x}$ $dy/dx = x^lnx xx (2lnx)/x$

$\frac{d y}{d x} = x^{ln x - 1} \times 2 ln x$ $dy/dx = x^(lnx - 1) xx 2lnx$

Hopefully this helps!

How do you find the derivative of y=xlnxy = x^(ln x)?