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. -v \( \quad t^{2}+2 t \quad t^{3}-3 \) 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.) \( \frac{d x}{d t}=2 t+2 \) (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 find \( z^{\prime}(t) \) using both methods, let's start with part (a). a. We have: \[ z = \frac{1}{x} + \frac{1}{y} = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] To differentiate \( z \) with respect to \( t \), we'll use the sum of derivatives: \[ \frac{dz}{dt} = \frac{d}{dt}\left(\frac{1}{t^2 + 2t}\right) + \frac{d}{dt}\left(\frac{1}{t^3 - 3}\right) \] Using the quotient rule for each term, we find: \[ \frac{dz}{dt} = -\frac{(2t + 2)}{(t^2 + 2t)^2} - \frac{(3t^2)}{(t^3 - 3)^2} \] Now for part (b). b. We start by identifying our intermediate derivatives: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] Given \( x = t^2 + 2t \), we found: \[ \frac{d x}{d t} = 2t + 2 \] Next: \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] For \( y = t^3 - 3 \), we need to find \( \frac{d y}{d t} \): \[ \frac{d y}{d t} = 3t^2 \] Now we can combine these using the chain rule: \[ z^{\prime}(t) = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t} \] Substituting in the expressions: \[ z^{\prime}(t) = -\frac{1}{(t^2 + 2t)^2} (2t + 2) - \frac{1}{(t^3 - 3)^2} (3t^2) \]