Application of Derivatives — Class 12 Maths Solution

We have,
$y = \cfrac{1}{{x - 1}},\,\,x \ne 1$ …(i)

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = \cfrac{{ - 1}}{{{{(x - 1)}^2}}}$

For tangents having slope $= - 1$, we must have-l $= \cfrac{{ - 1}}{{{{(x - 1)}^2}}}$
$\Rightarrow {(x - 1)^2} = 1 \Rightarrow x - 1 = \pm 1 \Rightarrow x = 1 \pm 1 = 2,0$

When $x = 2$, then from (i), $y = \cfrac{1}{{2 - 1}} = 1$ Therefore, The point is $(2,\;1)$.

Equation of the tangent at $(2,\;1)$ is $y - 1 = - 1(x - 2)$,
or $x + y - 3 = 0$

When $x = 0$, then from (i), $y = \cfrac{1}{{0 - 1}} = - 1$

Therefore, the point is $(0, - 1)$ .
Equation of tangent at $(0,\; - 1)$ is $y - ( - 1) = - 1(x - 0)$,
or $x + y + 1 = 0$

Therefore the required tangents are $x + y - 3 = 0$ and $x + y + 1 = 0.$