2. Find the slope of the tangent to the curve   \frac{x-1}{x-2} , x \neq 2 \: \: at\: \: x = 10

Answers (1)

Given curve is,

y = \frac{x-1}{x-2}
The slope of the tangent at x = 10 is given by
\left ( \frac{dy}{dx} \right )_{x=10}= \frac{(1)(x-2)-(1)(x-1)}{(x-2)^2} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}
 at x = 10
                       = \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}
hence, slope of tangent at x = 10 is \frac{-1}{64}

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