Find \( z^{\prime}(t) \), where \( z=\frac{1}{x}+\frac{1}{y}, x=t^{2}+2 t \), and \( y=t^{3}-3 \), in the following ways. a. Replace \( x \) and \( y \) to write \( z \) as a function of \( t \) and differentiate. b. Use the Chain Rule. b. To find the derivative using the Chain Rule, begin by finding the intermediate derivatives. \( \frac{\partial z}{\partial x}=-\frac{1}{t^{2}}+2 t \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=2 t+2 \) (Type an expression using \( t \) as the variable.) \( \frac{\partial z}{\partial y}=-\frac{1}{y^{2}} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d y}{d t}=\square \) (Type an expression using \( t \) as the variable.)

To begin with, we have \( z = \frac{1}{x} + \frac{1}{y} \) where \( x = t^2 + 2t \) and \( y = t^3 - 3 \). ### Part a: First, we substitute \( x \) and \( y \) into \( z \): \[ z = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] Now we will differentiate \( z \) with respect to \( t \): Using the formula for the derivative \( \frac{d}{dt}(u + v) = \frac{du}{dt} + \frac{dv}{dt} \): For the first term: \[ \frac{d}{dt}\left(\frac{1}{t^2 + 2t}\right) = -\frac{(2t + 2)}{(t^2 + 2t)^2} \] For the second term: \[ \frac{d}{dt}\left(\frac{1}{t^3 - 3}\right) = -\frac{(3t^2)}{(t^3 - 3)^2} \] Putting it all together, we find \( z'(t) \): \[ z'(t) = -\frac{(2t + 2)}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} \] ### Part b: Now, using the Chain Rule: The Chain Rule states that: \[ z' = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \] We have already determined: - \( \frac{\partial z}{\partial x} = -\frac{1}{x^2} \) - and \( \frac{dx}{dt} = 2t + 2 \) Next, we need to find \( \frac{\partial z}{\partial y} \) which we have: \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] Now we need to find \( \frac{dy}{dt} \): \[ y = t^3 - 3 \implies \frac{dy}{dt} = 3t^2 \] Thus, we summarize: - \( \frac{\partial z}{\partial x} = -\frac{1}{x^2} \) - \( \frac{dx}{dt} = 2t + 2 \) - \( \frac{\partial z}{\partial y} = -\frac{1}{y^2} \) - \( \frac{dy}{dt} = 3t^2 \) Now we can plug everything back into the Chain Rule formula: \[ z' = \left(-\frac{1}{x^2}\right)(2t + 2) + \left(-\frac{1}{y^2}\right)(3t^2) \] And there you have it! You can calculate specific values if needed, but this gives you the full derivative.