if x/x-y=log a/x-y, provethatdy/dx=2-x/ y 

We are given the equation:

\[
\frac{x}{x - y} = \log \frac{a}{x - y}
\]

We need to prove that:

\[
\frac{dy}{dx} = \frac{2 - x}{y}
\]

### Step 1: Differentiate the equation implicitly
Let's differentiate both sides of the equation with respect to \(x\).

- On the left-hand side, we have \( \frac{x}{x - y} \). This is a quotient, so we will apply the quotient rule:

\[
\frac{d}{dx} \left( \frac{x}{x - y} \right) = \frac{(x - y) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(x - y)}{(x - y)^2}
\]

This simplifies to:

\[
\frac{(x - y) \cdot 1 - x \cdot (1 - \frac{dy}{dx})}{(x - y)^2} = \frac{(x - y) - x + x \cdot \frac{dy}{dx}}{(x - y)^2}
\]

Which becomes:

\[
\frac{- y + x \cdot \frac{dy}{dx}}{(x - y)^2}
\]

- On the right-hand side, we have \( \log \frac{a}{x - y} \). By the chain rule, the derivative of the logarithmic function is:

\[
\frac{d}{dx} \left( \log \frac{a}{x - y} \right) = \frac{1}{\frac{a}{x - y}} \cdot \frac{d}{dx} \left( \frac{a}{x - y} \right)
\]

Since \(a\) is constant, this becomes:

\[
= \frac{1}{\frac{a}{x - y}} \cdot \frac{a \cdot \frac{dy}{dx}}{(x - y)^2} = \frac{1}{\frac{a}{x - y}} \cdot \frac{a \cdot (1 - \frac{dy}{dx})}{(x - y)^2}
Correctments above. Now.

If \( \frac{x}{x - y} = \log \frac{a}{x - y} \), prove that \( \frac{dy}{dx} = \frac{2 - x}{y} \)

This problem involves proving the derivative of an implicit logarithmic function. It uses implicit differentiation, the quotient rule, and logarithmic differentiation techniques to demonstrate that \( \frac{dy}{dx} = \frac{2 - x}{y} \). Suitable for university-level calculus students.

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