Question

For any positive constant \( a \), evaluate \[ \frac{dy}{dx}, \text{ where } y = a \frac{t+1}{t} \text{ and } x = (t + 1)^{\alpha}. \]

Hint:

When differentiating a quotient, use the quotient rule: \[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}, \text{where } y = \frac{u}{v}. \]

Answer 1

We are given: \[ y = a \frac{t+1}{t} \text{and} x = (t + 1)^{\alpha}. \] First, differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt} = a \left[ \frac{d}{dt} \left( \frac{t+1}{t} \right) \right]. \] Using the quotient rule: \[ \frac{dy}{dt} = a \left( \frac{t \cdot 1 - (t+1) \cdot 1}{t^2} \right) = a \left( \frac{t - (t+1)}{t^2} \right) = a \left( \frac{-1}{t^2} \right). \] Thus, \[ \frac{dy}{dt} = -\frac{a}{t^2}. \] Next, differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt} \left( (t+1)^{\alpha} \right) = \alpha (t + 1)^{\alpha-1}. \] Now, apply the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{a}{t^2}}{\alpha (t + 1)^{\alpha-1}}. \] Thus, \[ \frac{dy}{dx} = -\frac{a}{\alpha t^2 (t + 1)^{\alpha-1}}. \] Conclusion: The value of \( \frac{dy}{dx} \) is \[ \boxed{-\frac{a}{\alpha t^2 (t + 1)^{\alpha-1}}}. \]