# Find the slope of the tangent to the curve $$\normalsize y =\large { \frac{x-1}{x-2},} \normalsize x \neq 2\; at\;x = 10$$

$\begin{array}{1 1} (A)\;\large\frac{1}{64} \\ (B)\;\large\frac{-1}{64} \\ (C)\; 64 \\ (D)\;-64 \end{array}$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$=slope of the tangent to $y=f(x)$ at point $P$.
Step 1:
$y=\large\frac{x-1}{x-2}$
Differentiating w.r.t $x$ on both sides,by applying the quotient rule,
$\large\frac{dy}{dx}=\frac{(x-2)(1)-(x-1)(1)}{(x-2)^2}$
$\qquad=\large\frac{x-2-x+1}{(x-2)^2}$
$\qquad=\large\frac{-1}{(x-2)^2}$
Step 2:
$\large\frac{dy}{dx}$ at $x=10$
Substituting for $x$ we get,
$\large\frac{dy}{dx}_{(x=10)}=\large\frac{-1}{(10-2)^2}$
$\qquad\qquad=\large\frac{-1}{64}$
Hence the slope of the tangent to the curve $y=\large\frac{x-1}{x-2}$ is $\large\frac{-1}{64}$