Use implicit differentiation and the product rule 2. Find $$ \frac{d y}{d x} $$ implicitly for the following functions: $$ \text { a. } f(x)=x^{2}+2 x y^{4}=-2 $$ $$ \begin{array}{l}\text { b. } \ \qquad \begin{array}{l}4 \ 4\end{array} \ \qquad x^{5}=2 x^{3}\end{array} $$

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To find $$ \frac{d y}{d x} $$ implicitly for the given functions, we will apply implicit differentiation and the product rule where necessary.

### Part a: 
Given the equation:
$$
f(x) = x^{2} + 2xy^{4} = -2
$$

1. **Differentiate both sides with respect to $$ x $$**:
   $$
\frac{d}{dx}(x^{2}) + \frac{d}{dx}(2xy^{4}) = \frac{d}{dx}(-2)
   $$

2. **Differentiate each term**:
   - The derivative of $$ x^{2} $$ is $$ 2x $$.
   - For $$ 2xy^{4} $$, we use the product rule:
     $$
     \frac{d}{dx}(2xy^{4}) = 2\left( y^{4} \frac{d}{dx}(x) + x \frac{d}{dx}(y^{4}) \right) = 2\left( y^{4} + x \cdot 4y^{3} \frac{dy}{dx} \right)
     $$
   - The derivative of a constant $$-2$$ is $$0$$.

3. **Combine the derivatives**:
   $$
   2x + 2\left( y^{4} + 4xy^{3} \frac{dy}{dx} \right) = 0
   $$

4. **Simplify**:
   $$
   2x + 2y^{4} + 8xy^{3} \frac{dy}{dx} = 0
   $$

5. **Isolate $$ \frac{dy}{dx} $$**:
   $$
   8xy^{3} \frac{dy}{dx} = -2x - 2y^{4}
   $$
   $$
   \frac{dy}{dx} = \frac{-2x - 2y^{4}}{8xy^{3}} = \frac{-(x + y^{4})}{4xy^{3}}
   $$

### Part b:
Given the equation:
$$
x^{5} = 2x^{3}
$$

1. **Rearrange the equation**:
   $$
x^{5} - 2x^{3} = 0
   $$

2. **Differentiate both sides with respect to $$ x $$**:
   $$
\frac{d}{dx}(x^{5}) - \frac{d}{dx}(2x^{3}) = 0
   $$

3. **Differentiate each term**:
   - The derivative of $$ x^{5} $$ is $$ 5x^{4} $$.
   - The derivative of $$ 2x^{3} $$ is $$ 6x^{2} $$.

4. **Combine the derivatives**:
   $$
   5x^{4} - 6x^{2} = 0
   $$

5. **Factor out common terms**:
   $$
   x^{2}(5x^{2} - 6) = 0
   $$

6. **Set each factor to zero**:
   - $$ x^{2} = 0 $$ gives $$ x = 0 $$.
   - $$ 5x^{2} - 6 = 0 $$ gives $$ x^{2} = \frac{6}{5} $$ or $$ x = \pm \sqrt{\frac{6}{5}} $$.

Since there is no $$ y $$ term in this equation, we do not find $$ \frac{dy}{dx} $$ explicitly for part b, as it is not an implicit function of $$ y $$.

### Summary of Results:
- For part a:
$$
\frac{dy}{dx} = \frac{-(x + y^{4})}{4xy^{3}}
$$
- For part b, the solutions for $$ x $$ are:
$$
x = 0, \quad x = \sqrt{\frac{6}{5}}, \quad x = -\sqrt{\frac{6}{5}}
$$
For part a: $$ \frac{dy}{dx} = \frac{-(x + y^{4})}{4xy^{3}} $$ For part b: $$ x = 0, \quad x = \sqrt{\frac{6}{5}}, \quad x = -\sqrt{\frac{6}{5}} $$

Use implicit differentiation and the product rule 2. Find \( \frac{d y}{d x} \) implicitly for the following functions: \[ \text { a. } f(x)=x^{2}+2 x y^{4}=-2 \] \[ \begin{array}{l}\text { b. } \\ \qquad \begin{array}{l}4 \\ 4\end{array} \\ \qquad x^{5}=2 x^{3}\end{array} \]

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