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.)

Solve the differential equation by following steps: - step0: Solve using separation of variable: \(\frac{dz}{dt}=\frac{\frac{z}{x}\times \frac{x}{t}+\frac{z}{y}\times \frac{y}{t}}{t}\) - step1: Simplify: \(\frac{dz}{dt}=\frac{2z}{t^{2}}\) - step2: Rewrite the expression: \(\frac{dz}{dt}=\frac{2}{t^{2}}\times z\) - step3: Rewrite the expression: \(\frac{1}{z}\times \frac{dz}{dt}=\frac{2}{t^{2}}\times z\times \frac{1}{z}\) - step4: Multiply the terms: \(\frac{1}{z}\times \frac{dz}{dt}=\frac{2}{t^{2}}\) - step5: Transform the expression: \(\frac{1}{z}\times dz=\frac{2}{t^{2}}\times dt\) - step6: Integrate both sides of the equation: \(\int \frac{1}{z} dz=\int \frac{2}{t^{2}} dt\) - step7: Calculate: \(\ln{\left(z\right)}+C_{1}=\int \frac{2}{t^{2}} dt,C_{1} \in \mathbb{R}\) - step8: Calculate: \(\ln{\left(z\right)}+C_{1}=-\frac{2}{t}+C_{2},C_{1} \in \mathbb{R},C_{2} \in \mathbb{R}\) - step9: Rewrite the expression: \(\ln{\left(z\right)}=-\frac{2}{t}+C,C \in \mathbb{R}\) - step10: Calculate: \(z=e^{-\frac{2}{t}+C},C \in \mathbb{R}\) - step11: Rewrite the expression: \(z=Ce^{-\frac{2}{t}},C \in \mathbb{R}\) Solve the equation \( \frac{d z}{d t} = \frac{\frac{\text{d} z}{\text{d} x} \frac{\text{d} x}{\text{d} t} + \frac{\text{d} z}{\text{d} y} \frac{\text{d} y}{\text{d} t}}{\text{d} t} \). Solve the differential equation by following steps: - step0: Solve using separation of variable: \(\frac{dz}{dt}=\frac{\frac{z}{x}\times \frac{x}{t}+\frac{z}{y}\times \frac{y}{t}}{t}\) - step1: Simplify: \(\frac{dz}{dt}=\frac{2z}{t^{2}}\) - step2: Rewrite the expression: \(\frac{dz}{dt}=\frac{2}{t^{2}}\times z\) - step3: Rewrite the expression: \(\frac{1}{z}\times \frac{dz}{dt}=\frac{2}{t^{2}}\times z\times \frac{1}{z}\) - step4: Multiply the terms: \(\frac{1}{z}\times \frac{dz}{dt}=\frac{2}{t^{2}}\) - step5: Transform the expression: \(\frac{1}{z}\times dz=\frac{2}{t^{2}}\times dt\) - step6: Integrate both sides of the equation: \(\int \frac{1}{z} dz=\int \frac{2}{t^{2}} dt\) - step7: Calculate: \(\ln{\left(z\right)}+C_{1}=\int \frac{2}{t^{2}} dt,C_{1} \in \mathbb{R}\) - step8: Calculate: \(\ln{\left(z\right)}+C_{1}=-\frac{2}{t}+C_{2},C_{1} \in \mathbb{R},C_{2} \in \mathbb{R}\) - step9: Rewrite the expression: \(\ln{\left(z\right)}=-\frac{2}{t}+C,C \in \mathbb{R}\) - step10: Calculate: \(z=e^{-\frac{2}{t}+C},C \in \mathbb{R}\) - step11: Rewrite the expression: \(z=Ce^{-\frac{2}{t}},C \in \mathbb{R}\) Let's solve the problem step by step. ### Part a: Write \( z \) as a function of \( t \) and differentiate. Given: \[ z = \frac{1}{x} + \frac{1}{y}, \quad x = t^2 + 2t, \quad y = t^3 - 3 \] Substituting \( x \) and \( y \) into \( z \): \[ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] Now, we differentiate \( z(t) \) with respect to \( t \): \[ \frac{dz}{dt} = -\frac{1}{(t^2 + 2t)^2} \cdot \frac{d}{dt}(t^2 + 2t) - \frac{1}{(t^3 - 3)^2} \cdot \frac{d}{dt}(t^3 - 3) \] Calculating the derivatives: 1. For \( x = t^2 + 2t \): \[ \frac{dx}{dt} = 2t + 2 \] 2. For \( y = t^3 - 3 \): \[ \frac{dy}{dt} = 3t^2 \] Substituting these into the derivative of \( z \): \[ \frac{dz}{dt} = -\frac{1}{(t^2 + 2t)^2}(2t + 2) - \frac{1}{(t^3 - 3)^2}(3t^2) \] ### Part b: Use the Chain Rule. Using the Chain Rule: \[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \] We already have: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2}, \quad \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] Now substituting \( x \) and \( y \): \[ \frac{\partial z}{\partial x} = -\frac{1}{(t^2 + 2t)^2}, \quad \frac{\partial z}{\partial y} = -\frac{1}{(t^3 - 3)^2} \] Thus, we can write: \[ \frac{dz}{dt} = -\frac{1}{(t^2 + 2t)^2}(2t + 2) - \frac{1}{(t^3 - 3)^2}(3t^2) \] ### Summary of Results: - \( z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \) - \( \frac{dz}{dt} = -\frac{1}{(t^2 + 2t)^2}(2t + 2) - \frac{1}{(t^3 - 3)^2}(3t^2) \) This completes the differentiation of \( z \) with respect to \( t \) using both methods.