Question

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

Answer 1

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

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

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

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

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

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