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differentiation of
 ln y + (x/y) = c 
is
 (1/y) . (dy/dx) + ( ( y - (x (dy/dx) ) ) / y^2 ) 
why the dy/dx

To differentiate the equation \( \ln y + \frac{x}{y} = c \), we apply implicit differentiation. Let's break down the process step-by-step:

### Step 1: Differentiate both sides with respect to \( x \)
The equation we start with is:
\[
\ln y + \frac{x}{y} = c
\]
Here, \( c \) is a constant, and \( y \) is a function of \( x \), so we use implicit differentiation.

### Step 2: Differentiate \( \ln y \)
The derivative of \( \ln y \) with respect to \( x \) requires using the chain rule. Let’s calculate it:
\[
\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}
\]
This is because:
- The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \),
- And then, you multiply by the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} \).

Thus, the derivative of \( \ln y \) is:
\[
\frac{1}{y} \cdot \frac{dy}{dx}
\]

### Step 3: Differentiate \( \frac{x}{y} \)
Next, we differentiate the second term, \( \frac{x}{y} \). We need to use the quotient rule because it's a quotient of two functions, \( x \) and \( y \).

The quotient rule states:
\[
\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]
Here, \( u = x \) and \( v = y \).

- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = \frac{dy}{dx} \)

Applying the quotient rule:
\[
\frac{d}{dx}\left( \frac{x}{y} \right) = \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2}
\]

Thus, the derivative of \( \frac{x}{y} \) is:
\[
\frac{y - x \cdot \frac{dy}{dx}}{y^2}
\]

### Step 4: Differentiate the constant \( c \)
The derivative of the constant \( c \) is simply 0.

### Step 5: Combine the results
Now, we combine the derivatives:
\[
\frac{1}{y} \cdot \frac{dy}{dx} + \frac{y - x \cdot \frac{dy}{dx}}{y^2} = 0
\]

### Why is there \( \frac{dy}{dx} \)?
The term \( \frac{dy}{dx} \) appears because \( y \) is a function of \( x \). When we differentiate \( y \) with respect to \( x \), the chain rule requires us to account for the derivative of \( y \) with respect to \( x \), which is \( \frac{dy}{dx} \).

In both the \( \ln y \) term and the \( \frac{x}{y} \) term, differentiating \( y \) with respect to \( x \) introduces \( \frac{dy}{dx} \), which represents the rate of change of \( y \) with respect to \( x \).

---

Would you like a more detailed step on how to solve for \( \frac{dy}{dx} \), or do you have any specific questions?

Here are 5 related questions:
1. How does implicit differentiation work with trigonometric functions?
2. Can you explain the chain rule in more detail for multivariable functions?
3. How do we differentiate exponential functions implicitly?
4. What is the quotient rule, and when should it be used in differentiation?
5. How does implicit differentiation differ from regular differentiation?

**Tip**: Whenever you're dealing with logarithms and quotient functions, always consider applying the chain and quotient rules for implicit differentiation.

differentiation of ln y + (x/y) = c is (1/y) * (dy/dx) + ( ( y - (x * (dy/dx) ) ) / y^2 ) why the dy/dx

Learn how to differentiate the equation ln y + (x/y) = c using implicit differentiation. This problem covers key concepts such as logarithmic differentiation, the chain rule, and the quotient rule. Suitable for students at the undergraduate level studying calculus.

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