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 $\frac{d}{dx}(ax^n)=nax^{n - 1}$, if $y=-5x^{2}-15$, then $\frac{dy}{dx}=-10x$.

Step2: Use the chain - rule

The chain - rule states that $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$. We want to find $\frac{dt}{dy}$, and we know that $\frac{dt}{dy}=\frac{1}{\frac{dy}{dt}}$. Since $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$, then $\frac{dt}{dy}=\frac{1}{\frac{dy}{dx}\cdot\frac{dx}{dt}}$.

Step3: Evaluate $\frac{dy}{dx}$ at $x = - 1$

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

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

We know that $\frac{dx}{dt}=3$. Then $\frac{dt}{dy}=\frac{1}{\frac{dy}{dx}\cdot\frac{dx}{dt}}$. Substitute $\frac{dy}{dx}=10$ and $\frac{dx}{dt}=3$ into the formula, we have $\frac{dt}{dy}=\frac{1}{10\times3}=\frac{1}{30}$.

Answer:

$\frac{1}{30}$