If ` y= Ke^(x//(x-a)) ` then ` (dy)/(dx)` is equal to

Options

A. `a (x-a)^2`
B. `-(ay)/((x-a)^2)`
C. `a^2 (x-a)^2)`
D. None of these

Answer

To find the derivative of the function \( y = K e^{\frac{x}{x-a}} \) with respect to \( x \), we will use the chain rule and the quotient rule. Let's go through the solution step by step.

### Step 1: Identify the function
We have:
\[ y = K e^{\frac{x}{x-a}} \]
Here, \( K \) is a constant.

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we will apply the chain rule. The derivative of \( e^u \) is \( e^u \frac{du}{dx} \), where \( u = \frac{x}{x-a} \).

### Step 3: Differentiate the exponent
Now we need to differentiate \( u = \frac{x}{x-a} \). We will use the quotient rule for this:
\[ u = \frac{x}{x-a} \]
Let \( U = x \) and \( V = x - a \). According to the quotient rule:
\[ \frac{du}{dx} = \frac{V \frac{dU}{dx} - U \frac{dV}{dx}}{V^2} \]
Calculating the derivatives:
- \( \frac{dU}{dx} = 1 \)
- \( \frac{dV}{dx} = 1 \)

Now substituting into the quotient rule:
\[ \frac{du}{dx} = \frac{(x-a)(1) - x(1)}{(x-a)^2} = \frac{x-a - x}{(x-a)^2} = \frac{-a}{(x-a)^2} \]

### Step 4: Apply the chain rule
Now substituting \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = K e^{\frac{x}{x-a}} \cdot \frac{du}{dx} = K e^{\frac{x}{x-a}} \cdot \frac{-a}{(x-a)^2} \]

### Step 5: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = \frac{-a K e^{\frac{x}{x-a}}}{(x-a)^2} \]

### Final Result
The derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{-a K e^{\frac{x}{x-a}}}{(x-a)^2} \]

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