se implicit differentiation to determine $$ \frac{d y}{d x} $$ for the equation $$ y e^{4 x}+5 y-1=0 $$

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To find $$ \frac{d y}{d x} $$ using implicit differentiation for the equation

$$
y e^{4x} + 5y - 1 = 0,
$$

we will differentiate both sides of the equation with respect to $$ x $$.

### Step 1: Differentiate both sides

Differentiating the left side:

1. For the term $$ y e^{4x} $$, we will use the product rule:
   - Let $$ u = y $$ and $$ v = e^{4x} $$.
   - Then, $$ \frac{du}{dx} = \frac{dy}{dx} $$ and $$ \frac{dv}{dx} = 4e^{4x} $$.
   - By the product rule: 
   $$
   \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} = y(4e^{4x}) + e^{4x} \frac{dy}{dx}.
   $$

2. For the term $$ 5y $$, we differentiate it as:
   $$
   \frac{d}{dx}(5y) = 5 \frac{dy}{dx}.
   $$

3. The derivative of the constant $$ -1 $$ is $$ 0 $$.

Putting it all together, we have:

$$
y(4e^{4x}) + e^{4x} \frac{dy}{dx} + 5 \frac{dy}{dx} = 0.
$$

### Step 2: Collect terms involving $$ \frac{dy}{dx} $$

Now, we can rearrange the equation to isolate $$ \frac{dy}{dx} $$:

$$
e^{4x} \frac{dy}{dx} + 5 \frac{dy}{dx} = -y(4e^{4x}).
$$

Factoring out $$ \frac{dy}{dx} $$:

$$
\frac{dy}{dx}(e^{4x} + 5) = -y(4e^{4x}).
$$

### Step 3: Solve for $$ \frac{dy}{dx} $$

Now, we can solve for $$ \frac{dy}{dx} $$:

$$
\frac{dy}{dx} = \frac{-y(4e^{4x})}{e^{4x} + 5}.
$$

Thus, the derivative $$ \frac{dy}{dx} $$ is:

$$
\frac{dy}{dx} = \frac{-4ye^{4x}}{e^{4x} + 5}.
$$
$$ \frac{dy}{dx} = \frac{-4ye^{4x}}{e^{4x} + 5} $$

se implicit differentiation to determine \( \frac{d y}{d x} \) for the equation \( y e^{4 x}+5 y-1=0 \)

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