Derivative of x^(lnx)^ln(lnx)?

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Derivative of x^(lnx)^ln(lnx)?

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Answer 1 · 2018-06-14T16:26:48

$x^{{(ln x)}^{ln (ln x)} - 1} {(ln x)}^{ln (ln x)} (2 ln (ln x) + 1)$ $x^((lnx)^(ln(lnx))-1)(lnx)^ln(lnx)(2ln(lnx)+1)$

Explanation:

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

Take the natural log of both sides:

$ln y = ln (x^{{(ln x)}^{ln (ln x)}}) = {(ln x)}^{ln (ln x)} (ln x) = {(ln x)}^{ln (ln x) + 1}$ $lny=ln(x^((lnx)^(ln(lnx))))=(lnx)^ln(lnx)(lnx)=(lnx)^(ln(lnx)+1)$

Take the natural log once more:

$ln (ln y) = ln ({(ln x)}^{ln (ln x) + 1}) = (ln (ln x) + 1) (ln (ln x))$ $ln(lny)=ln((lnx)^(ln(lnx)+1))=(ln(lnx)+1)(ln(lnx))$

Now take the derivative of both sides. Use the chain rule on the left and product and chain rules on the right.

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

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

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

Solving for the derivative:

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

Substituting in $y$ $y$ and $ln y$ $lny$ as defined and previously derived:

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

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

Answer 2 · 2018-06-14T16:56:58

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

Explanation:

Let,

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

For simplicity we take,

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

So, $y = x^{N}$ $y=x^N$

Taking log ,we get

$ln y = {ln x}^{N} = N \cdot ln x$ $lny=lnx^N=N*lnx$ , where , $N = {(ln x)}^{ln (ln x)}$ $N=(lnx)^(ln(lnx))$

$=>lny=(lnx)^(ln(lnx))*lnx...to(A)$

Again taking log,we get

$ln(lny)=ln((lnx)^(ln(lnx))*lnx)$

$=>ln(lny)=ln((lnx)^(ln(lnx)))+ln(lnx)$

$=>ln(lny)=ln(lnx)(ln(lnx)+ln(lnx)$

$=>ln(lny)=ln(lnx)[ln(lnx)+1]$

Diff. w. r. t. $x$ , $"using "color(blue)"product and Chain Rule :"$

$1/(lny)*1/y(dy)/(dx)$ = $ln(lnx)[1/lnx*1/x]+[ln(lnx)+1]xx1/lnx1/x$

$=>1/(lny)*1/y(dy)/(dx)$ = $[1/lnx*1/x]{ln(lnx)+ln(lnx)+1}$

$=>1/(ylny)(dy)/(dx)$ = $[1/(xlnx)]{2ln(lnx)+1}$

$=>(dy)/(dx)=(ylny)/(xlnx){2ln(lnx)+1}$

OR. from $(A)$ , we can put ,

$color(blue)(lny=(lnx)^(ln(lnx))*lnx)$

So,

$(dy)/(dx)=(y)/(xlnx)xxcolor(blue)((lnx)^(ln(lnx)) *lnx) {2ln(lnx)+1}$

$=>(dy)/(dx)=(y)/(x)xx(lnx)^(ln(lnx)){2ln(lnx)+1}$

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Derivative of x^(lnx)^ln(lnx)?

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