Find the slope of the tangent to the curve `y=(x-3)/(x-5), x ne 5" at " x=10`.

Answer

To find the slope of the tangent to the curve \( y = \frac{x - 3}{x - 5} \) at \( x = 10 \), we will follow these steps:

### Step 1: Differentiate the function using the Quotient Rule

The given function is:
\[ y = \frac{x - 3}{x - 5} \]
To find the slope of the tangent line, we need to compute the derivative \( \frac{dy}{dx} \). We will use the Quotient Rule, which states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( u = x - 3 \) and \( v = x - 5 \).

### Step 2: Compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)

Calculating the derivatives:
\[ \frac{du}{dx} = 1 \quad \text{(since the derivative of \( x - 3 \) is 1)} \]
\[ \frac{dv}{dx} = 1 \quad \text{(since the derivative of \( x - 5 \) is 1)} \]

### Step 3: Apply the Quotient Rule

Now substituting into the Quotient Rule:
\[ \frac{dy}{dx} = \frac{(x - 5)(1) - (x - 3)(1)}{(x - 5)^2} \]
Simplifying the numerator:
\[ = \frac{x - 5 - (x - 3)}{(x - 5)^2} = \frac{x - 5 - x + 3}{(x - 5)^2} = \frac{-2}{(x - 5)^2} \]

### Step 4: Evaluate the derivative at \( x = 10 \)

Now we need to find the slope of the tangent at \( x = 10 \):
\[ \frac{dy}{dx} \bigg|_{x=10} = \frac{-2}{(10 - 5)^2} = \frac{-2}{5^2} = \frac{-2}{25} \]

### Conclusion

Thus, the slope of the tangent to the curve at \( x = 10 \) is:
\[ \frac{-2}{25} \]

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