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Calculus: Differentiability and Curve Properties - Class XI

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If slope of tangent of curve $y=\dfrac{x}{b-x}$ at $(1,1)$ be $2$ then $b=$

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$2$

💡 Explanation:

Given,

$y=\dfrac{x}{b-x}$

Now,

$\dfrac{dy}{dx}=\dfrac{(b-x)1+x.(-1)}{(b-x)^2}$

or, $\dfrac{dy}{dx}=\dfrac{b}{(b-x)^2}$.

Now,

$\left.\dfrac{dy}{dx}\right| _{(1,1)}=\dfrac{b}{(b-1)^2}$.

According to the problem,

$\dfrac{b}{(b-1)^2}=2$

or, $b=2(b^2-2b+1)$
or, $2b^2-5b+2=0$

or, $(2b-1)(b-2)=0$

or, $b=\dfrac{1}{2}, 2$.