We are given the equation:

$\tan(4x + y) = 4x$

We need to differentiate this implicitly with respect to $x$, and then solve for $\frac{dy}{dx}$. We will treat $y$ as a function of $x$ (i.e., $y = y(x)$).

Step 1: Differentiate both sides with respect to $x$

To differentiate $\tan(4x + y)$, we will apply the chain rule. Recall that the derivative of $\tan(u)$ with respect to $u$ is $\sec^2(u)$, and the derivative of $u = 4x + y(x)$ will require the sum rule and chain rule for $y(x)$.

$\frac{d}{dx}[\tan(4x + y)] = \sec^2(4x + y) \cdot \frac{d}{dx}(4x + y)$

Now, differentiate $4x + y$:

$\frac{d}{dx}(4x + y) = 4 + \frac{dy}{dx}$

So, the derivative of the left-hand side is:

$\sec^2(4x + y) \cdot (4 + \frac{dy}{dx})$

Step 2: Differentiate the right-hand side

The derivative of $4x$ with respect to $x$ is simply $4$.

Thus, we now have:

$\sec^2(4x + y) \cdot (4 + \frac{dy}{dx}) = 4$

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

Now, let's solve this equation for $\frac{dy}{dx}$:

$\sec^2(4x + y) \cdot (4 + \frac{dy}{dx}) = 4$

First, divide both sides by $\sec^2(4x + y)$:

$4 + \frac{dy}{dx} = \frac{4}{\sec^2(4x + y)}$

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

$\frac{dy}{dx} = \frac{4}{\sec^2(4x + y)} - 4$

Using the identity $\sec^2(\theta) = 1 + \tan^2(\theta)$, we can simplify the denominator:

$\frac{dy}{dx} = \frac{4}{1 + \tan^2(4x + y)} - 4$

Step 4: Final expression

So, the derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = \frac{4}{1 + \tan^2(4x + y)} - 4$

Do you need further clarification or detailed steps on any part?

Here are some related questions you might find interesting:

1. How do you simplify the derivative if $y$ were a constant?
2. Can you apply implicit differentiation to find higher-order derivatives?
3. What is the geometric interpretation of the derivative of the tangent function?
4. How would the result change if we replaced $\tan$ with $\sin$?
5. What is the role of the chain rule in solving implicit derivatives?

Tip: When dealing with implicit differentiation, always remember to differentiate each term separately, keeping in mind that $y$ is a function of $x$.