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

Explanation:

Step1: Find the first - derivative (y') using the product rule ((uv)^\prime = u^\prime v+uv^\prime)

Let (u = x) and (v=(3x + 4)^4). Then (u^\prime=1) and (v^\prime = 4(3x + 4)^3\times3=12(3x + 4)^3) (using the chain rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)), where (f(u)=u^4), (u = 3x+4), (f^\prime(u) = 4u^3), (g^\prime(x)=3)).

[ \begin{align*} y^\prime&=(x)^\prime(3x + 4)^4+x\cdot[(3x + 4)^4]^\prime\ &=(3x + 4)^4+x\cdot12(3x + 4)^3\ &=(3x + 4)^3[(3x + 4)+12x]\ &=(3x + 4)^3(15x + 4) \end{align*} ]

Step2: Find the second - derivative (y'') using the product rule again

Let (u=(3x + 4)^3) and (v=(15x + 4)). Then (u^\prime=3(3x + 4)^2\times3 = 9(3x + 4)^2) (chain rule) and (v^\prime=15)

[ \begin{align*} y''&=(3x + 4)^3\times15+(15x + 4)\times9(3x + 4)^2\ &=3(3x + 4)^2[5(3x + 4)+3(15x + 4)]\ &=3(3x + 4)^2(15x+20 + 45x+12)\ &=3(3x + 4)^2(60x + 32)\ &=6(3x + 4)^2(30x + 16)\ &=6(3x + 4)^2\times2(15x + 8)\ &=12(3x + 4)^2(15x + 8) \end{align*} ]

Answer:

(y''=12(3x + 4)^2(15x + 8))