# How do you differentiate x^x?

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There are two ways we can find the derivative of x^x. It's important to notice that this function is neither a power function of the form x^k nor an exponential function of the form b^x, so we can't use the differentiation formulas for either of these cases directly.

(i) Let y=x^x, and take logarithms of both sides of this equation: ln(y)=ln(x^x). Using properties of logarithmic functions, we can rewrite this as ln(y)=x.ln(x). Then differentiating both sides with respect to x, and using the chain rule on the LHS and product rule on the RHS, gives 1/y.dy/dx=ln(x)+1. Rearranging, we have dy/dx=y.(ln(x)+1). That is, dy/dx=x^x(ln(x)+1).

(ii) Write x^x=e^(ln(x^x))=e^(x.ln(x)), using the properties of the exponential and logarithmic functions. Now, d/dx(x.ln(x))=ln(x)+1 by the product rule. Hence, d/dx(e^(x.ln(x)))=(ln(x)+1).(e^(x.ln(x)) by the chain rule, and using the fact that the derivative of e^[f(x)]=f'(x).e^[f(x)] for any differentiable function f(x). Finally, rewriting e^(x.ln(x)) as x^x gives d/dx(x^x)=x^x.(ln(x)+1), as with the first method.

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