\( 1 \leftarrow \) 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. 0 of 1 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 \( t \). \( z(t)=\square \)

To express \( z \) as a function of \( t \), we first need to substitute \( x \) and \( y \) with their respective expressions in terms of \( t \): 1. Given \( x = t^2 + 2t \) 2. Given \( y = t^3 - 3 \) Now, substituting these back into the equation for \( z \): \[ z(t) = \frac{1}{x} + \frac{1}{y} = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] Thus, we have: \[ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} \] This is the expression for \( z \) as a function of \( t \).