Question

find y. y = x(3x + 2)^4 y =

Explanation:

Step1: Use the product rule \( (uv)^\prime=u^\prime v + uv^\prime \)

Let \( u = x \), \( u^\prime=1 \); \( v=(3x + 2)^4 \).
To find \( v^\prime \), use the chain rule. Let \( t = 3x+2 \), \( v=t^4 \). Then \( \frac{dv}{dt}=4t^3 \), \( \frac{dt}{dx}=3 \), so \( v^\prime=\frac{dv}{dt}\cdot\frac{dt}{dx}=12(3x + 2)^3 \).
By the product rule, \( y^\prime=(3x + 2)^4+12x(3x + 2)^3=(3x + 2)^3(3x + 2+12x)=(3x + 2)^3(15x + 2) \).

Step2: Use the product rule again for \( y^\prime=(3x + 2)^3(15x + 2) \)

Let \( u=(3x + 2)^3 \), \( v = 15x+2 \).
For \( u^\prime \), use the chain rule. Let \( t = 3x+2 \), \( u=t^3 \), \( \frac{du}{dt}=3t^2 \), \( \frac{dt}{dx}=3 \), so \( u^\prime = 9(3x + 2)^2 \). And \( v^\prime=15 \).
By the product rule \( y^{\prime\prime}=9(3x + 2)^2(15x + 2)+15(3x + 2)^3 \)
Factor out \( 3(3x + 2)^2 \):
\( y^{\prime\prime}=3(3x + 2)^2[3(15x + 2)+5(3x + 2)] \)
Expand the brackets inside: \( 45x+6 + 15x+10=60x + 16 \)
So \( y^{\prime\prime}=3(3x + 2)^2(60x + 16)=12(3x + 2)^2(15x + 4) \)

Answer:

\(12(3x + 2)^2(15x + 4)\)