for the function y=(4x + 3)(2x^2+7x - 1), find $\frac{d^{2}y}{dx^{2}}$. answer: $\frac{d^{2}y}{dx^{2}}=$

Question

Accepted Answer

# Explanation:
## Step1: Expand the function
$$
\begin{align*}
y&=(4x + 3)(2x^{2}+7x - 1)\
&=4x(2x^{2}+7x - 1)+3(2x^{2}+7x - 1)\
&=8x^{3}+28x^{2}-4x + 6x^{2}+21x - 3\
&=8x^{3}+34x^{2}+17x - 3
\end{align*}
$$
## Step2: Find the first - derivative
Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have:
$$
\begin{align*}
\frac{dy}{dx}&=\frac{d}{dx}(8x^{3}+34x^{2}+17x - 3)\
&=8\times3x^{2}+34\times2x+17\
&=24x^{2}+68x + 17
\end{align*}
$$
## Step3: Find the second - derivative
Again, using the power rule:
$$
\begin{align*}
\frac{d^{2}y}{dx^{2}}&=\frac{d}{dx}(24x^{2}+68x + 17)\
&=24\times2x+68\
&=48x + 68
\end{align*}
$$
# Answer:
$48x + 68$

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QUESTION IMAGE

Question

Explanation:

Step1: Expand the function

Step2: Find the first - derivative

Step3: Find the second - derivative

Answer: