Question

what is the value of $\frac{d}{dx}(2x^{4}+x^{3}+3x^{2})$ at $x = - 1$? choose 1 answer: a -4 b -3 c -11 d 4

Explanation:

Step1: Differentiate the function

Use the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.
For $y = 2x^{4}+x^{3}+3x^{2}$, $\frac{dy}{dx}=2\times4x^{3}+3x^{2}+3\times2x=8x^{3}+3x^{2}+6x$.

Step2: Substitute $x=-1$

Substitute $x = - 1$ into $\frac{dy}{dx}$.
$\frac{dy}{dx}\big|_{x=-1}=8\times(-1)^{3}+3\times(-1)^{2}+6\times(-1)$.
$=8\times(-1)+3\times1 - 6$.
$=-8 + 3-6=-11$.

Answer:

C. - 11