what is the value of $\frac{d}{dx}(2x^{2}-6x + 5)$ at $x=-4$? choose 1 answer: a 61 b -14 c -10 d -22
what is the value of $\frac{d}{dx}(2x^{2}-6x + 5)$ at $x=-4$? choose 1 answer: a 61 b -14 c -10 d -22
Answer
Explanation:
Step1: Differentiate the function
Use the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$. For $y = 2x^2-6x + 5$, $\frac{dy}{dx}=\frac{d}{dx}(2x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)$. So, $\frac{dy}{dx}=4x-6$.
Step2: Substitute $x=-4$
Substitute $x = - 4$ into $\frac{dy}{dx}$. We get $\frac{dy}{dx}\big|_{x=-4}=4\times(-4)-6$. Calculate $4\times(-4)-6=-16 - 6=-22$.
Answer:
D. - 22