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Find Derivative of y = x^{ x}

A calculus tutorial on how to find the first derivative of y = x^{ x} for x > 0.

Note that the function defined by y = x ^{ x} is neither a power function of the form x^{ k} nor an exponential function of the form b^{ x} and the formulas of Differentiation of these functions cannot be used. We need to find another method to find the first derivative of the above function.
If y = x ^{ x} and x > 0 then ln y = ln (x^{ x})
Use properties of logarithmic functions to expand the right side of the above equation as follows. ln y = x ln x We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right. y '(1 / y) = ln x + x(1 / x) = ln x + 1 , where y ' = dy/dx Multiply both sides by y y ' = (ln x + 1)y Substitute y by x ^{ x} to obtainy ' = (ln x + 1)x ^{ x}Exercise: Find the first derivative of y = x^{x - 2}Answer to Above Exercise: y ' = x^{ x - 3} (x ln x + x - 2)
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