Question

Find dy/dx using log differentiation. y = x^(x-1)

Answer 1

\[
y = x^{x-1} \\
\log(y) = \log(x)^{x-1} = (x-1)\log(x) \\
\text{Implicit Differentiation on both sides:} \\
\frac{dy}{y} = ((x-1)\frac 1x + \log(x))dx\\
\frac{dy}{dx} = y * ((x-1)\frac 1x + \log(x))\\
\frac{dy}{dx} = x^{x-1} * ((x-1)\frac 1x + \log(x))\\
\]

Answer 2

\[
\frac{dy}{dx} = x^{x-1} * ((x-1)\frac 1x + \log(x))\\
\frac{dy}{dx} = x^{x-1} * ((1-\frac 1x) + \log(x))\\
\frac{dy}{dx} = x^{x-1} - x^{x-2} + x^{x-1} \log(x)\\
\]

Answer 3

Thank you!