How do you find the first and second derivative of ${(ln x)}^{ln x}$ $(lnx)^lnx$ ?

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How do you find the first and second derivative of ${(ln x)}^{ln x}$ $(lnx)^lnx$ ?

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Answer 1 · 2016-11-11T23:06:03

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

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

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

Let $y = ln u$ $y= lnu$ and $u = ln x$ $u = lnx$ .

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

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

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

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

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

The second derivative can be found using the product rule to differentiate the upper term and the quotient rule to differentiate the entire expression.

You can do the algebra.

Hopefully this helps!

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How do you find the first and second derivative of ${(ln x)}^{ln x}$ $(lnx)^lnx$ ?

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