Questions: Find dy/dt for each pair of functions. y=x^2-8x, x=t^2+2 dy/dt=

Solution Steps

To find \(\frac{dy}{dt}\) for the given pair of functions, we need to use the chain rule. The chain rule states that if \(y\) is a function of \(x\) and \(x\) is a function of \(t\), then \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\).

1. First, find \(\frac{dy}{dx}\) by differentiating \(y = x^2 - 8x\) with respect to \(x\).
2. Next, find \(\frac{dx}{dt}\) by differentiating \(x = t^2 + 2\) with respect to \(t\).
3. Finally, multiply the results from steps 1 and 2 to get \(\frac{dy}{dt}\).

Given: \[ y = x^2 - 8x \]

We need to find \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{d}{dx}(x^2 - 8x) \]

Using the power rule and the constant multiple rule: \[ \frac{dy}{dx} = 2x - 8 \]

Given: \[ x = t^2 + 2 \]

We need to find \(\frac{dx}{dt}\): \[ \frac{dx}{dt} = \frac{d}{dt}(t^2 + 2) \]

Using the power rule: \[ \frac{dx}{dt} = 2t \]

We need to find \(\frac{dy}{dt}\). Using the chain rule: \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \]

Substitute \(\frac{dy}{dx}\) and \(\frac{dx}{dt}\) from the previous steps: \[ \frac{dy}{dt} = (2x - 8) \cdot 2t \]

Given \( x = t^2 + 2 \), substitute this into the expression: \[ \frac{dy}{dt} = (2(t^2 + 2) - 8) \cdot 2t \]

Simplify the expression inside the parentheses: \[ 2(t^2 + 2) - 8 = 2t^2 + 4 - 8 = 2t^2 - 4 \]

So: \[ \frac{dy}{dt} = (2t^2 - 4) \cdot 2t \]

Distribute \(2t\): \[ \frac{dy}{dt} = 2t \cdot (2t^2 - 4) = 4t^3 - 8t \]

Final Answer