Derivative of x^x [x to the power x] | d/dx(x^x)

The derivative of x^x is equal to x^x(1+ln x). That is, d/dx(x^x)=x^x(1+ln x). This can be proved by the logarithmic differentiation. Here, ln=log_e. In this post, we will learn how to find the derivative of x to the x.

Derivative of x^x Formula

The formula of x^x derivative is given by

d/dx(x^x)=x^x(1+ln x)

Derivative of x^x Proof

Now, we will give a proof of the derivative of x^x, that is, d/dx(x^x) = x^x(1+ln x).

Proof:

Let y=x^x.

Taking the natural logarithms both sides, we have

ln y = ln x^x

⇒ ln y =x ln x.

Taking d/dx on both sides, we have

$\dfrac{d}{dx}(\ln y)=\dfrac{d}{dx}{x \ln x}$

Now, applying the product rule of derivatives, it follows that

$\dfrac{1}{y} \dfrac{dy}{dx}$ $=x\dfrac{d}{dx}{\ln x}+\ln x\dfrac{dx}{dx}$

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=x \cdot \dfrac{1}{x}+\ln x$ as the derivative of lnx is 1/x and d/dx(x)=1.

⇒ $\dfrac{dy}{dx}=y(1+\ln x)$

⇒ $\dfrac{dy}{dx}=x^x(1+\ln x)$ as y=x^x.

Thus, the derivative of x^x (x to the x) is equal to x^x(1+ln x) and this is obtained by the logarithmic differentiation.

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