The problem is to find $\frac{dy}{dx}$ given the equation $\sin x + \frac{\log(2y)}{4y} = 5$. We need to use implicit differentiation.

1. Problem Description

The problem is to find $\frac{dy}{dx}$ given the equation $\sin x + \frac{\log(2y)}{4y} = 5$. We need to use implicit differentiation.

2. Solution Steps

Given the equation $\sin x + \frac{\log(2y)}{4y} = 5$.

First, differentiate both sides with respect to $x$:

$\frac{d}{dx} (\sin x) + \frac{d}{dx} \left( \frac{\log(2y)}{4y} \right) = \frac{d}{dx} (5)$

We know that $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(5) = 0$.

So, $\cos x + \frac{d}{dx} \left( \frac{\log(2y)}{4y} \right) = 0$.

Now we need to find the derivative of $\frac{\log(2y)}{4y}$ with respect to $x$. We'll use the quotient rule:

$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Let $u = \log(2y)$ and $v = 4y$.

Then $\frac{du}{dx} = \frac{1}{2y} \cdot 2 \cdot \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$

and $\frac{dv}{dx} = 4 \frac{dy}{dx}$.

So,

$\frac{d}{dx} \left( \frac{\log(2y)}{4y} \right) = \frac{4y \cdot \frac{1}{y} \frac{dy}{dx} - \log(2y) \cdot 4 \frac{dy}{dx}}{(4y)^2} = \frac{4 \frac{dy}{dx} - 4 \log(2y) \frac{dy}{dx}}{16y^2} = \frac{4 (1 - \log(2y)) \frac{dy}{dx}}{16y^2} = \frac{(1 - \log(2y)) \frac{dy}{dx}}{4y^2}$

Now, plug this back into the original differentiated equation:

$\cos x + \frac{(1 - \log(2y))}{4y^2} \frac{dy}{dx} = 0$

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

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

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

Multiply both the numerator and denominator by -1:

$\frac{dy}{dx} = -\frac{4y^2 \cos x}{1 - \log(2y)}$.

The given solution is:

$\frac{dy}{dx} = -\frac{3 y^2 \cos x}{4 - \log(2y)}$

The correct derivative of $log(2y)$ with respect to x is $\frac{1}{2y} * 2 * \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$.

The correct derivative of $4y$ with respect to x is $4 \frac{dy}{dx}$.

Using quotient rule on $\frac{log(2y)}{4y}$ will give:

$\frac{4y*\frac{1}{y}\frac{dy}{dx} - log(2y)*4\frac{dy}{dx}}{(4y)^2} = \frac{4\frac{dy}{dx} - 4log(2y)\frac{dy}{dx}}{16y^2} = \frac{(1-log(2y))\frac{dy}{dx}}{4y^2}$.

Therefore $\frac{dy}{dx} = - \frac{4y^2cos(x)}{1 - log(2y)}$.

The question must be a copy error. The equation for $\frac{dy}{dx}$ is most probably:

$\frac{dy}{dx} = -\frac{4 y^2 \cos x}{4 - \log(2y)}$.

This gives:

$cos x + \frac{dy}{dx}\frac{1 - log(2y)}{4y^2} = 0$

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

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

If we wish to obtain $-\frac{3 y^2 \cos x}{4 - \log(2y)}$, there is no algebraic way from the given equation.

3. Final Answer

$\frac{dy}{dx} = - \frac{4y^2 \cos x}{1 - \log(2y)}$. Assuming there was a typo in the question, a more likely solution would be: $\frac{dy}{dx} = -\frac{4 y^2 \cos x}{4 - \log(2y)}$