Question

question: if (y = - 5x^{2}-15) and (\frac{dx}{dt}=3), find (\frac{dy}{dt}) at (x = - 1). provide your answer below: (\frac{dy}{dt}=square)

Explanation:

Step1: Differentiate y with respect to x

Using the power - rule, if $y=-5x^{2}-15$, then $\frac{dy}{dx}=-10x$.

Step2: Evaluate $\frac{dy}{dx}$ at $x = - 1$

Substitute $x=-1$ into $\frac{dy}{dx}$, we get $\frac{dy}{dx}\big|_{x = - 1}=-10\times(-1)=10$.

Step3: Use the chain - rule $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$

We know that $\frac{dx}{dt}=3$. Rearranging for $\frac{dy}{dt}$ gives $\frac{dy}{dt}=10\times3 = 30$. Then, since $\frac{dy}{dt}=30$, we can find $\frac{dt}{dy}=\frac{1}{\frac{dy}{dt}}$.

Step4: Calculate $\frac{dt}{dy}$

$\frac{dt}{dy}=\frac{1}{30}$.

Answer:

$\frac{1}{30}$