\( 1 \leftarrow \) 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 find \( z^{\prime}(t) \) where \( z = \frac{1}{x} + \frac{1}{y} \), \( x = t^2 + 2t \), and \( y = t^3 - 3 \), we will follow the two methods outlined. ### a. Write \( z \) as a function of \( t \) and differentiate. We have already expressed \( z \) as a function of \( t \): \[ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] Now, we will differentiate \( z(t) \) with respect to \( t \). Using the quotient rule for differentiation, where \( \frac{d}{dt} \left( \frac{1}{u} \right) = -\frac{u'}{u^2} \): 1. For the first term \( \frac{1}{t^2 + 2t} \): - Let \( u = t^2 + 2t \) - Then \( u' = 2t + 2 \) Thus, \[ \frac{d}{dt} \left( \frac{1}{u} \right) = -\frac{2t + 2}{(t^2 + 2t)^2} \] 2. For the second term \( \frac{1}{t^3 - 3} \): - Let \( v = t^3 - 3 \) - Then \( v' = 3t^2 \) Thus, \[ \frac{d}{dt} \left( \frac{1}{v} \right) = -\frac{3t^2}{(t^3 - 3)^2} \] Now, combining these results, we have: \[ z'(t) = -\frac{2t + 2}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} \] ### b. Use the Chain Rule. To find \( z' \) using the Chain Rule, we need to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), and then multiply by the derivatives of \( x \) and \( y \) with respect to \( t \). 1. **Finding \( \frac{\partial z}{\partial x} \)**: \[ z = \frac{1}{x} + \frac{1}{y} \] Thus, \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] 2. **Finding \( \frac{\partial z}{\partial y} \)**: \[ \frac{\partial z}{\partial y} = -\frac{1}{y^2} \] 3. **Finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)**: - For \( x = t^2 + 2t \): \[ \frac{dx}{dt} = 2t + 2 \] - For \( y = t^3 - 3 \): \[ \frac{dy}{dt} = 3t^2 \] Now, applying the Chain Rule: \[ z'(t) = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \] Substituting the derivatives we found: \[ z'(t) = -\frac{1}{x^2} (2t + 2) - \frac{1}{y^2} (3t^2) \] Thus, we have: \[ \frac{\partial z}{\partial x} = -\frac{1}{x^2} \] This completes the solution for both parts.