Landyn Allison

2023-03-22

What is the derivative of ${x}^{\mathrm{ln}x}$?

chicka152m0w3

The derivative of ${x}^{\mathrm{ln}x}$ is $\left[\frac{2\cdot y\cdot \left(\mathrm{ln}x\right)\cdot \left({x}^{\mathrm{ln}x}\right)}{x}\right]$
Step 1
let $y={x}^{\mathrm{ln}x}$
There are no rules that we can apply to easily differentiate this equation, so we just have to mess with it until we find an answer.
By taking the natural log of both sides, we change the equation. We can do it as long as we remember that this will be a completely new equation:
$\mathrm{ln}y=\mathrm{ln}\left({x}^{\mathrm{ln}x}\right)$
$\mathrm{ln}y=\left(\mathrm{ln}x\right)\left(\mathrm{ln}x\right)$
Differentiate both sides:
$\left(\frac{dy}{dx}\right)\cdot \left(\frac{1}{y}\right)=\left(\mathrm{ln}x\right)\left(\frac{1}{x}\right)+\left(\frac{1}{x}\right)\left(\mathrm{ln}x\right)$
$\left(\frac{dy}{dx}\right)=\frac{2\cdot y\cdot \mathrm{ln}x}{x}$
Then, we're done messing with that equation. Let's go back to the original problem:
$y={x}^{\mathrm{ln}x}$
We can rewrite this as $y={e}^{\mathrm{ln}\left({x}^{\mathrm{ln}x}\right)}$ because e to the power of a natural log of some number is that same number.
$y={e}^{\mathrm{ln}\left({x}^{\mathrm{ln}x}\right)}$
Next, let's differentiate this using the exponent rule:
$\frac{dy}{dx}=\frac{d}{dx}\left[\mathrm{ln}\left({x}^{\mathrm{ln}x}\right)\right]\cdot \left[{e}^{\mathrm{ln}\left({x}^{\mathrm{ln}x}\right)}\right]$
Conveniently, we already found the first term above, so we can easily simplify this.
$\frac{dy}{dx}=\left[\frac{2\cdot y\cdot \mathrm{ln}x}{x}\right]\cdot \left[{x}^{\mathrm{ln}x}\right]$
$\frac{dy}{dx}=\frac{2\cdot y\cdot \left(\mathrm{ln}x\right)\cdot \left({x}^{\mathrm{ln}x}\right)}{x}$

Engerlingagis

Let $y={x}^{\mathrm{ln}x}$
Take Natural logs of both sides
$\mathrm{ln}y={\mathrm{ln}x}^{\mathrm{ln}x}$
$\therefore \mathrm{ln}y=\left(\mathrm{ln}x\right)\left(\mathrm{ln}x\right)$
$\therefore \mathrm{ln}y=\left({\mathrm{ln}}^{2}x\right)$
Differentiate (implicitly) wrt x on the LHS and apply the chain rule to the RHS:
$\therefore \frac{1}{y}\frac{dy}{dx}=\left(2\mathrm{ln}x\right)\frac{1}{x}$
$\therefore \frac{dy}{dx}=\frac{2y\mathrm{ln}x}{x}$
$\therefore \frac{dy}{dx}=\frac{\left(2{x}^{\mathrm{ln}x}\right)\mathrm{ln}x}{x}$
$\therefore \frac{dy}{dx}=\left(\left(2{x}^{\mathrm{ln}x}\right)\mathrm{ln}x\right){x}^{-1}$
$\therefore \frac{dy}{dx}=\left(2{x}^{\left(\mathrm{ln}x\right)-1}\right)\mathrm{ln}x$

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