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

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Answer 1 · 2015-11-01T10:12:31

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

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

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

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

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

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

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