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Derivative of $$$x^{x}$$$
Related calculator: Derivative Calculator
Solution
Use the formula $$$f^{g{\left(x \right)}}{\left(x \right)} = e^{g{\left(x \right)} \ln\left(f{\left(x \right)}\right)}$$$ with $$$f{\left(x \right)} = x$$$ and $$$g{\left(x \right)} = x$$$ to rewrite the complex expression:
$${\color{red}\left(\frac{d}{dx} \left(x^{x}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)}$$The function $$$e^{x \ln\left(x\right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = e^{u}$$$ and $$$g{\left(x \right)} = x \ln\left(x\right)$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}$$The derivative of the exponential is $$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$Return to the old variable:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = e^{{\color{red}\left(x \ln\left(x\right)\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = x^{x} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$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 $$$g{\left(x \right)} = \ln\left(x\right)$$$:
$$x^{x} {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = x^{x} {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$x^{x} \left(x \frac{d}{dx} \left(\ln\left(x\right)\right) + \ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}\right) = x^{x} \left(x \frac{d}{dx} \left(\ln\left(x\right)\right) + \ln\left(x\right) {\color{red}\left(1\right)}\right)$$The derivative of the natural logarithm is $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$x^{x} \left(x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right)\right) = x^{x} \left(x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right)\right)$$Thus, $$$\frac{d}{dx} \left(x^{x}\right) = x^{x} \left(\ln\left(x\right) + 1\right)$$$.
Answer
$$$\frac{d}{dx} \left(x^{x}\right) = x^{x} \left(\ln\left(x\right) + 1\right)$$$A