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Derivative of $$$x^{x}$$$

The calculator will find the derivative of $$$x^{x}$$$ using the logarithmic differentiation, with steps shown.

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Find $$$\frac{d}{dx} \left(x^{x}\right)$$$.

Solution

Let $$$H{\left(x \right)} = x^{x}$$$.

Take the logarithm of both sides: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(x^{x}\right)$$$.

Rewrite the RHS using the properties of logarithms: $$$\ln\left(H{\left(x \right)}\right) = x \ln\left(x\right)$$$.

Differentiate separately both sides of the equation: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(x \ln\left(x\right)\right)$$$.

Differentiate the LHS of the equation.

The function $$$\ln\left(H{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$

Return to the old variable:

$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$

Thus, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.

Differentiate the RHS of the equation.

Apply the product rule $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ with $$$f{\left(x \right)} = x$$$ and $$$g{\left(x \right)} = \ln\left(x\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:

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

Thus, $$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$.

Hence, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \ln\left(x\right) + 1$$$.

Therefore, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\ln\left(x\right) + 1\right) H{\left(x \right)} = x^{x} \left(\ln\left(x\right) + 1\right)$$$.

Answer

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

Related Note: Logarithmic Differentiation