Question

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

Answer 1

$\frac{du\left(x\right)}{v\left(x\right)}=\frac{\frac{d\left(u\right(x\left)\right)}{dx}v\left(x\right)-u\left(x\right)\frac{d\left(v\right(x\left)\right)}{dx}}{{v\left(x\right)}^2}$

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

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

at $x=10$ we get $\frac{-1}{64}$