$$ 1 \leftarrow $$ 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.)

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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.
To find $$ z^{\prime}(t) $$, where $$ z = \frac{1}{x} + \frac{1}{y} $$, $$ x = t^2 + 2t $$, and $$ y = t^3 - 3 $$, we can differentiate $$ z $$ with respect to $$ t $$ using two methods: **a. Direct Differentiation:** $$ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 3} $$ $$ z'(t) = -\frac{2t + 2}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} $$ **b. Using the Chain Rule:** $$ \frac{\partial z}{\partial x} = -\frac{1}{x^2} $$ $$ \frac{\partial z}{\partial y} = -\frac{1}{y^2} $$ $$ \frac{dx}{dt} = 2t + 2, \quad \frac{dy}{dt} = 3t^2 $$ $$ z'(t) = -\frac{1}{x^2} (2t + 2) - \frac{1}{y^2} (3t^2) $$ Thus, $$ z^{\prime}(t) $$ is: $$ z^{\prime}(t) = -\frac{2t + 2}{(t^2 + 2t)^2} - \frac{3t^2}{(t^3 - 3)^2} $$

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