Question

Question #5c268

Answer 1

Logarithmic Differentiation

`f(x)=(x-1)^x`

by taking the natural log, and `ln x^r=r ln x`,

`=> ln[f(x)]=ln(x-1)^x=x ln(x-1)`

by implicitly differentiating with respect to `x`,

`=> {f'(x)}/{f(x)}=1 cdot ln(x-1)+x cdot 1/{x-1}=ln(x-1)+x/{x-1}`

by multiplying by `f(x)=(x-1)^x`,

`=>f'(x)=[ln(x-1)+x/{x-1}] (x-1)^x`

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I hope that this was helpful.