Use logarithmic differentiation to find the derivative of the function y = x^x

Solution:

\(y = x^{x}\)

Taking log on both sides

\(log y = logx^{x}\)

\(log y = xlogx\)     (logxⁿ = nlogx)

Differentiating both sides we have

\(\frac{\mathrm{d} logy}{\mathrm{d} x} = \frac{\mathrm{d} (xlogx)}{\mathrm{d} x}\)

\(\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x} = x\frac{\mathrm{d} logx}{\mathrm{d} x} + (logx)(\frac{\mathrm{d} x}{\mathrm{d} x})\)

\(\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x} = x(1/x) + logx\)

\(\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x} = 1 + logx\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = y(1 + logx)\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = (1 + logx)x^{x}\)

Use logarithmic differentiation to find the derivative of the function y = x^x

Summary:

The derivative of the function y = x^x using logarithmic definition is \((1 + logx)x^{x}\)