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

1 Answer

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)#