Question

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

Answer 1

Given curve is $y=\frac{x-1}{x-2}$

On differentiating, we get

$\frac{dy}{dx}=\frac{(x-2)\frac{d}{dx}(x-1)-(x-1)\frac{d}{dx}(x-2)}{(x-2{)}^2}$(Using quotient rule)

$\Rightarrow \frac{dy}{dx}=\frac{(x-2)\times 1-(x-1)\times 1}{(x-2{)}^2}=\frac{x-2-x+1}{(x-2{)}^2}=\frac{-1}{(x-2{)}^2}$

$\therefore$ Slope of tangent at x=10 is ${\left(\frac{dy}{dx}\right)}_{x=10}$ $=\frac{-1}{(10-2{)}^2}=-\frac{1}{8^2}=-\frac{1}{64}$

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