question\nfor the following equation, evaluate $\frac{dy}{dx}$ when $x = 1$.\n$y=-3x^{4}-x^{2}$
question\nfor the following equation, evaluate $\frac{dy}{dx}$ when $x = 1$.\n$y=-3x^{4}-x^{2}$
Answer
Explanation:
Step1: Differentiate using power rule
The power - rule states that if $y = ax^n$, then $\frac{dy}{dx}=nax^{n - 1}$. For $y=-3x^{4}-x^{2}$, we have: $\frac{dy}{dx}=\frac{d}{dx}(-3x^{4})+\frac{d}{dx}(-x^{2})$. Applying the power - rule: $\frac{d}{dx}(-3x^{4})=-3\times4x^{4 - 1}=-12x^{3}$ and $\frac{d}{dx}(-x^{2})=-2x^{2 - 1}=-2x$. So, $\frac{dy}{dx}=-12x^{3}-2x$.
Step2: Substitute $x = 1$
Substitute $x = 1$ into $\frac{dy}{dx}=-12x^{3}-2x$. $\frac{dy}{dx}\big|{x = 1}=-12\times(1)^{3}-2\times(1)$. $\frac{dy}{dx}\big|{x = 1}=-12 - 2=-14$.
Answer:
$-14$