\( 1 \leftarrow \) 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. \( \frac{\partial z}{\partial x}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

To differentiate \( z \) with respect to \( x \) using \( z=\frac{1}{x}+\frac{1}{y} \), we can apply the derivative rules: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] And since \( y \) is a function of \( t \), we also need to include the derivative of \( y \) with respect to \( x \) when applying the Chain Rule, which is often computed as \( \frac{\partial z}{\partial y} \cdot \frac{dy}{dx} \): \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] Now, using the Chain Rule, you can relate the derivatives of \( z \) to the derivatives of \( t \): To get \( z'(t) \) using the Chain Rule, you’ll need: 1. \( \frac{dx}{dt} = 2t + 2 \) 2. \( \frac{dy}{dt} = 3t^2 \) Combined, the Chain Rule gives: \[ z'(t) = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \] Inserting the expressions we found: \[ z'(t) = \left(-\frac{1}{x^2}\right)(2t+2) + \left(-\frac{1}{y^2}\right)(3t^2) \] This leads to the complete form for \( z'(t) \): Combine all components to derive the expression for \( z'(t) \).