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. a. Write \( z \) as a function of \( t \). \( z(t)=\frac{1}{t^{2}+2 t}+\frac{1}{t^{3}-3} \) b. To find the derivative using the Chain Rule, begin by finding the intermediate derivatives. \( \frac{\partial z}{\partial x}=-\frac{1}{x^{2}} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=\square \) (Type an expression using \( t \) as the variable.)

To find \( z' = z^{\prime}(t) \) using the Chain Rule, we first calculate the derivatives of \( x \) and \( y \) with respect to \( t \). 1. The derivative of \( x \) with respect to \( t \) is: \[ \frac{d x}{d t} = \frac{d}{dt}(t^{2} + 2t) = 2t + 2. \] 2. The derivative of \( y \) with respect to \( t \) is: \[ \frac{d y}{d t} = \frac{d}{dt}(t^{3} - 3) = 3t^{2}. \] Now we can proceed to calculate \( z^{\prime}(t) \) using the expression for \( z \): \[ z = \frac{1}{x} + \frac{1}{y}. \] 3. Using the Chain Rule, we have: \[ z^{\prime}(t) = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t}. \] From earlier, we already found \( \frac{\partial z}{\partial x} = -\frac{1}{x^{2}} \). Now we need \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial y} = -\frac{1}{y^{2}}. \] Putting it all together: \[ z^{\prime}(t) = -\frac{1}{x^{2}}(2t + 2) - \frac{1}{y^{2}}(3t^{2}). \] Thus, the fully calculated derivatives are: - \( \frac{d x}{d t} = 2t + 2 \) - \( \frac{d y}{d t} = 3t^{2} \)