Question

Can someone show me how to differentiate y=6 lnx? I'm sure it's chain rule, but I'm struggling with how to set it up.

Answer 1

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

Answer 2

Also,
\( \dfrac {d}{dx} ku = k \dfrac{d}{dx}u \)

Answer 3

\(y = 6 \ln x\)

\(y' = 6 \dfrac{d}{dx} \ln x\)

\(y' = 6 \cdot\dfrac{1}{x} \cdot \dfrac{d}{dx}x\)

\(y' = 6 \cdot\dfrac{1}{x} \cdot 1\)

\(y' = \dfrac{6}{x}\)