Question

How can we prove ln(x^r)=(r*lnx) using derivatives?

Answer 1

\(\dfrac{d}{dx} \ln u = \dfrac{1}{u} \dfrac{du}{dx}\)

\(\dfrac{d}{dx} \ln x^r = \dfrac{1}{x^r} \dfrac{d}{dx} x^r\)

\(\dfrac{d}{dx} \ln x^r = \dfrac{1}{x^r} rx^{r - 1}\)

\(\dfrac{d}{dx} \ln x^r = x^{-r} rx^{r - 1}\)

\(\dfrac{d}{dx} \ln x^r = rx^{- 1}\)

\(\dfrac{d}{dx} \ln x^r = \dfrac{r}{ x} \)

Now integrate both sides to get

\( \ln x^r = r \ln x\)

Answer 2

this ln x^r = rln x so this comes from property of logarithms