Can someone tell me where my error is in finding the derivative of y=x^(lnx)?

I'm using the rule $\frac{d}{\mathrm{dx}} {a}^{u} = {a}^{u} \ln a \frac{\mathrm{du}}{\mathrm{dx}}$. $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\ln x} \cdot \ln x \cdot \frac{1}{x}$ $= \frac{{x}^{\ln x} \ln x}{x}$ The correct answer is $= \frac{2 {x}^{\ln x} \ln x}{x}$.

Jan 16, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{\ln} x \ln x}{x}$

Explanation:

I'm pretty fond of logarithmic differentiation. We try to take the derivative of both sides:

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

$\ln y = \ln x \ln x$

Now use implicit differentiation and the product rule.

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

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

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

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

Or

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{\ln} x \ln x}{x}$

As required.

Hopefully this helps!

Jan 16, 2018

I got $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \ln \left(x\right) {x}^{\ln} x}{x}$ which can be rewritten as $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {x}^{\ln} x \ln \left(x\right)}{x}$

Explanation:

You've used the wrong derivative technique!

$\frac{d}{\mathrm{dx}} {a}^{u} = {a}^{u} \ln a \frac{\mathrm{du}}{\mathrm{dx}}$

This can only be used when $a$ is a constant such as:

${5}^{x} \to \frac{d}{\mathrm{dx}} = {5}^{x} \ln \left(5\right)$

But we have an $x$ as the base!

So we must use a special technique which involves talking the natural log $\left(\ln\right)$ of both sides and using implicit differentiation

Given: $y = {x}^{\ln} x$

Take $\ln$ of both sides

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

Since color(blue)(ln(x^a)=aln(x)

$\ln \left(y\right) = \ln \left(x\right) \cdot \ln \left(x\right)$

Differentiate both sides W.R.T $x$ (The right side requires the product rule)

For the left side: $\frac{d}{\mathrm{dx}} \left(\ln \left(y\right)\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y}$

For the right side:

Product rule: $\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

Let $f \left(x\right) = \ln \left(x\right)$ and $g \left(x\right) = \ln \left(x\right)$

Thus $f ' \left(x\right) = \frac{1}{x}$ and $g ' \left(x\right) = \frac{1}{x}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{1}{y} = \ln \frac{x}{x} + \ln \frac{x}{x}$

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

Multiply both sides by $y$

dy/dx=(2ln(x))/x*color(red)(y

We want to rewrite everything in terms of $x$ but we have this color(red)(y in the way. However, recall that $\textcolor{red}{y}$ is defined as $\textcolor{red}{y = {x}^{\ln} x}$

dy/dx=(2ln(x))/x*color(red)(x^lnx

Rewriting we get:

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

color(red)("This can also be rewritten as" color(red)(dy/dx=(2x^lnxln(x))/x color(red)("which correlates to the correct answer you provided."