Find `dy/dx if log(4x)+log(16x)=4y`.

Answer

To find \(\frac{dy}{dx}\) for the equation \(\log(4x) + \log(16x) = 4y\), we can follow these steps:

### Step 1: Combine the logarithmic terms
Using the property of logarithms that states \(\log(a) + \log(b) = \log(ab)\), we can combine the left side of the equation:

\[ \log(4x) + \log(16x) = \log(4x \cdot 16x) = \log(64x^2) \]

So, we rewrite the equation as:

\[ \log(64x^2) = 4y \]

### Step 2: Differentiate both sides with respect to \(x\)
Now we differentiate both sides of the equation. Remember that we will use the chain rule on the left side:

\[ \frac{d}{dx}[\log(64x^2)] = \frac{d}{dx}[4y] \]

Using the chain rule, the derivative of \(\log(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\):

\[ \frac{1}{64x^2} \cdot \frac{d}{dx}[64x^2] = 4 \frac{dy}{dx} \]

### Step 3: Differentiate \(64x^2\)
Now we differentiate \(64x^2\):

\[ \frac{d}{dx}[64x^2] = 128x \]

So, substituting this back, we have:

\[ \frac{1}{64x^2} \cdot 128x = 4 \frac{dy}{dx} \]

### Step 4: Simplify the equation
Now we simplify the left side:

\[ \frac{128x}{64x^2} = \frac{2}{x} \]

Thus, we have:

\[ \frac{2}{x} = 4 \frac{dy}{dx} \]

### Step 5: Solve for \(\frac{dy}{dx}\)
To isolate \(\frac{dy}{dx}\), we divide both sides by 4:

\[ \frac{dy}{dx} = \frac{2}{4x} = \frac{1}{2x} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{1}{2x} \]

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