How do you differentiate f(x)=4x ln(3x + 2) using the chain rule?

1 Answer

4 (ln(3x + 2) + x cdot \frac{1}{3x + 2} cdot 3)

Explanation:

\frac{d}{dx}[4x ln(3x + 2)] // constant out

= 4 \frac{d}{dx}[x ln(3x + 2)] //product rule

= 4 (\frac{d}{dx}[x] ln(3x + 2) + x \frac{d}{dx}[ln(3x + 2)]) // u = 3x + 2

= 4 (ln(3x + 2) + x \frac{d}{du}[ln u] \frac{d}{dx}[u])

= 4 (ln(3x + 2) + x cdot \frac{1}{u} cdot 3)