The derivative of x^logx (x to the power logx) is equal to 2 logx x^{logx-1}. Here we find the differentiation of x^{logx} by using the product rule of derivatives together with logarithmic differentiation.

The product rule of derivatives states the following:

$\dfrac{d}{dx}$(f(x)g(x)) = f(x) $\dfrac{d}{dx}$(g(x)) + g(x) $\dfrac{d}{dx}$(f(x)) **…(I)**

## Find the Derivative of x^{logx}

Answer: The derivative of x^{log x} is 2x^{logx-1} logx. |

We find the derivative of x^{logx} as described below. Here we have considered natural logarithms, that is, logx = log_{e} x.

Put y=x^{logx} **…(II)**

To find $\frac{dy}{dx}$, we use the method of logarithmic differentiation. Taking logarithms on both sides of (II) we get that

log_{e} y = log_{e} x^{logx}

⇒ log y = logx log x, here we have used the logarithm rule: log a^{m} = m log a.

Differentiating both sides we obtain that

$\dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{d}{dx}$(logx logx)

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ = logx $\dfrac{d}{dx}$(logx) + logx $\dfrac{d}{dx}$(logx) using the above product rule (I) of derivatives.

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ = logx ⋅ $\dfrac{1}{x}$ + logx ⋅ $\dfrac{1}{x}$

⇒ $\dfrac{dy}{dx}$ = y [$\dfrac{2 \log x}{x}$]

⇒ $\dfrac{dy}{dx}$ = x^{logx} [$\dfrac{2 \log x}{x}$] as y=x^{logx}.

⇒ $\dfrac{dy}{dx}$ = 2 logx x^{logx-1}

So the derivative of x^{logx} is equal to 2 logx x^{logx-1}.

Related Derivatives:

## FAQs

**Q1: What is the derivative of x ^{logx}?**

**Answer:** The derivative of x^{logx} (x raised to the logx) is equal to 2x^{logx-1} logx.

**Q2: If y= x ^{logx} , then find dy/dx.**

**Answer:** If y=x^{logx}, then dy/dx= 2 logx x^{logx-1}.