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
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
Answer
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