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

Jun 14, 2018

${x}^{{\left(\ln x\right)}^{\ln \left(\ln x\right)} - 1} {\left(\ln x\right)}^{\ln} \left(\ln x\right) \left(2 \ln \left(\ln x\right) + 1\right)$

Explanation:

$y = {x}^{{\left(\ln x\right)}^{\ln \left(\ln x\right)}}$

Take the natural log of both sides:

$\ln y = \ln \left({x}^{{\left(\ln x\right)}^{\ln \left(\ln x\right)}}\right) = {\left(\ln x\right)}^{\ln} \left(\ln x\right) \left(\ln x\right) = {\left(\ln x\right)}^{\ln \left(\ln x\right) + 1}$

Take the natural log once more:

$\ln \left(\ln y\right) = \ln \left({\left(\ln x\right)}^{\ln \left(\ln x\right) + 1}\right) = \left(\ln \left(\ln x\right) + 1\right) \left(\ln \left(\ln x\right)\right)$

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

$\frac{1}{\ln} y \left(\frac{d}{\mathrm{dx}} \ln y\right) = \left(\frac{d}{\mathrm{dx}} \left(\ln \left(\ln x\right) + 1\right)\right) \ln \left(\ln x\right) + \left(\ln \left(\ln x\right) + 1\right) \left(\frac{d}{\mathrm{dx}} \ln \left(\ln x\right)\right)$

$\frac{1}{\ln} y \left(\frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{\ln} x \frac{d}{\mathrm{dx}} \ln x\right) \ln \left(\ln x\right) + \left(\ln \left(\ln x\right) + 1\right) \left(\frac{1}{\ln} x \frac{d}{\mathrm{dx}} \ln x\right)$

$\frac{1}{y \ln y} \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \frac{\ln x}{x \ln x} + \frac{\ln \left(\ln x\right) + 1}{x \ln x}$

Solving for the derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \ln y \left(2 \ln \left(\ln x\right) + 1\right)}{x \ln x}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{{\left(\ln x\right)}^{\ln \left(\ln x\right)}} {\left(\ln x\right)}^{\ln \left(\ln x\right) + 1} \left(2 \ln \left(\ln x\right) + 1\right)}{x \ln x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{{\left(\ln x\right)}^{\ln \left(\ln x\right)} - 1} {\left(\ln x\right)}^{\ln} \left(\ln x\right) \left(2 \ln \left(\ln x\right) + 1\right)$

Jun 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x} \times {\left(\ln x\right)}^{\ln \left(\ln x\right)} \left\{2 \ln \left(\ln x\right) + 1\right\}$

Explanation:

Let,

y=x^((lnx)^(ln(lnx))

For simplicity we take,

$N = {\left(\ln x\right)}^{\ln \left(\ln x\right)}$

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

Taking log ,we get

$\ln y = \ln {x}^{N} = N \cdot \ln x$ , where , $N = {\left(\ln x\right)}^{\ln \left(\ln x\right)}$

$\implies \ln y = {\left(\ln x\right)}^{\ln \left(\ln x\right)} \cdot \ln x \ldots \to \left(A\right)$

Again taking log,we get

$\ln \left(\ln y\right) = \ln \left({\left(\ln x\right)}^{\ln \left(\ln x\right)} \cdot \ln x\right)$

$\implies \ln \left(\ln y\right) = \ln \left({\left(\ln x\right)}^{\ln \left(\ln x\right)}\right) + \ln \left(\ln x\right)$

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

$\implies \ln \left(\ln y\right) = \ln \left(\ln x\right) \left[\ln \left(\ln x\right) + 1\right]$

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

$\frac{1}{\ln y} \cdot \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$=$\ln \left(\ln x\right) \left[\frac{1}{\ln} x \cdot \frac{1}{x}\right] + \left[\ln \left(\ln x\right) + 1\right] \times \frac{1}{\ln} x \frac{1}{x}$

$\implies \frac{1}{\ln y} \cdot \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$=$\left[\frac{1}{\ln} x \cdot \frac{1}{x}\right] \left\{\ln \left(\ln x\right) + \ln \left(\ln x\right) + 1\right\}$

$\implies \frac{1}{y \ln y} \frac{\mathrm{dy}}{\mathrm{dx}}$=$\left[\frac{1}{x \ln x}\right] \left\{2 \ln \left(\ln x\right) + 1\right\}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \ln y}{x \ln x} \left\{2 \ln \left(\ln x\right) + 1\right\}$

OR. from $\left(A\right)$ , we can put ,

$\textcolor{b l u e}{\ln y = {\left(\ln x\right)}^{\ln \left(\ln x\right)} \cdot \ln x}$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x \ln x} \times \textcolor{b l u e}{{\left(\ln x\right)}^{\ln \left(\ln x\right)} \cdot \ln x} \left\{2 \ln \left(\ln x\right) + 1\right\}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x} \times {\left(\ln x\right)}^{\ln \left(\ln x\right)} \left\{2 \ln \left(\ln x\right) + 1\right\}$