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. \( \frac{\partial z}{\partial x}=-\frac{1}{x^{2}} \) (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.) \( \frac{d y}{d t}=3 t^{2} \) (Type an expression using \( t \) as the variable.) Using either method, \( z^{\prime}(t)=\square \). (Type an expression using \( t \) as the variable.)

**Method (a): Substitute \( x \) and \( y \) as functions of \( t \)** We have \[ z(t) = \frac{1}{x} + \frac{1}{y} = \frac{1}{t^2+2t} + \frac{1}{t^3-3}. \] Differentiate term-by-term: 1. For the first term: \[ \frac{d}{dt}\left(\frac{1}{t^2+2t}\right) = \frac{d}{dt}\left((t^2+2t)^{-1}\right) = -1\cdot (t^2+2t)^{-2} \cdot \frac{d}{dt}(t^2+2t). \] Since \[ \frac{d}{dt}(t^2+2t) = 2t+2, \] we get: \[ \frac{d}{dt}\left(\frac{1}{t^2+2t}\right) = -\frac{2t+2}{(t^2+2t)^2}. \] 2. For the second term: \[ \frac{d}{dt}\left(\frac{1}{t^3-3}\right) = \frac{d}{dt}\left((t^3-3)^{-1}\right) = -1\cdot (t^3-3)^{-2} \cdot \frac{d}{dt}(t^3-3). \] Since \[ \frac{d}{dt}(t^3-3) = 3t^2, \] we have: \[ \frac{d}{dt}\left(\frac{1}{t^3-3}\right) = -\frac{3t^2}{(t^3-3)^2}. \] Combining both results: \[ z'(t) = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-3)^2}. \] --- **Method (b): Use the Chain Rule** Given: \[ z = \frac{1}{x} + \frac{1}{y}, \] with \[ x=t^2+2t,\quad y=t^3-3, \] and the provided partial derivatives: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2},\quad \frac{\partial z}{\partial y} = -\frac{1}{y^2}, \] and the derivatives of \( x \) and \( y \) with respect to \( t \): \[ \frac{dx}{dt} = 2t+2,\quad \frac{dy}{dt} = 3t^2. \] By the Chain Rule: \[ z'(t) = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} = -\frac{1}{x^2}(2t+2) - \frac{1}{y^2}(3t^2). \] Substitute \( x=t^2+2t \) and \( y=t^3-3 \): \[ z'(t) = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-3)^2}. \] --- **Final Answer:** \[ z'(t) = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-3)^2}. \]