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Question

Mathematics Question on Applications of Derivatives

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

Answer

The given curve is y=x−1x−2\frac{x-1}{x-2}.

dydx\frac{dy}{dx}=(x−2)(1)−(x−1)(1)(x−2)2\frac {(x-2)(1)-(x-1)(1)}{(x-2)^2}

=x−2−x+1(x−2)2\frac {x-2-x+1}{(x-2)^2}=-−1(x−2)2\frac{-1}{(x-2)^2}

Thus, the slope of the tangent at x = 10 is given by,

(\frac{\text{dy}}{\text{dx}}) \bigg]_{x=10}$$= \frac{-1}{(x-2)^2}\bigg] _{ x=10}=-−1(10−2)2\frac{-1}{(10-2)^2}=−164\frac{-1}{64}

Hence, the slope of the tangent at x = 10 is −164\frac{-1}{64}