Question

compute the following.
$left.\frac{d^{2}}{dx^{2}}(2x^{3}-x^{2}+7x - 1)
ight|_{x = 4}$
$left.\frac{d^{2}}{dx^{2}}(2x^{3}-x^{2}+7x - 1)
ight|_{x = 4}=square$ (simplify your answer.)

Explanation:

Step1: Find the first - derivative

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $y = 2x^{3}-x^{2}+7x - 1$, we have $\frac{dy}{dx}=2\times3x^{2}-2x + 7=6x^{2}-2x + 7$.

Step2: Find the second - derivative

Differentiate $\frac{dy}{dx}=6x^{2}-2x + 7$ again using the power rule. $\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(6x^{2}-2x + 7)=12x-2$.

Step3: Evaluate at $x = 4$

Substitute $x = 4$ into $\frac{d^{2}y}{dx^{2}}$. $\frac{d^{2}y}{dx^{2}}\big|_{x = 4}=12\times4-2=48 - 2=46$.

Answer:

$46$