Question

please answer this question​

Answer 1

$\rm \longrightarrow \: y = \dfrac{x + 2}{ {x}^{2} - 3}$

Differentiating both sides :-

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{d \bigg(\dfrac{x + 2}{ {x}^{2} - 3} \bigg)}{dx}$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{({x}^{2} - 3) \dfrac{d(x + 2)}{dx} \: - \: (x + 2)\dfrac{d({x}^{2} - 3)}{dx} }{ {({x}^{2} - 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{({x}^{2} - 3) - (x + 2)(2x)}{ {({x}^{2} - 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{({x}^{2} - 3) - (2 {x}^{2} + 4x)}{ {({x}^{2} - 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{{x}^{2} - 3- 2 {x}^{2} - 4x}{ {({x}^{2} - 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{ - 3- {x}^{2} - 4x}{ {({x}^{2} - 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{ - ( 3 + {x}^{2} + 4x)}{ {({x}^{2} - 3)}^{2} }$

Now we have to find derivative at x = 0

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{ - ( 3 + 0 + 0)}{ {(0- 3)}^{2} }$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{ - 3}{ 9}$

$\rm \longrightarrow \: \dfrac{dy}{dx} = \dfrac{ - 1}{ 3}$

Answer 2

Step-by-step explanation:

$\large\bf{\underline{\underline{We\ have:-}}}$

$\bf : \implies {y\ =\ \dfrac{x\ +\ 2}{x^2\ -\ 3}}$

$\bf{\underline{\underline{By\ differentiating\ both\ sides:-}}}$

$\bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{d \bigg (\dfrac{x\ +\ 2}{x^2\ -\ 3} \bigg)}{dx}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{(x^2\ -\ 3)\ \dfrac{d(x\ +\ 2)}{dx}\ -\ (x\ +\ 2)\ \dfrac{d(x^2\ -\ 3)}{dx}}{(x^2\ -\ 3)^2}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{(x^2\ -\ 3)\ -\ (x\ +\ 2)\ (2x)}{(x^2\ -\ 3)^2}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{(x^2\ -\ 3)\ -\ (2x^2\ +\ 4x)}{(x^2\ -\ 3)^2}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{x^2\ -\ 3\ -\ 2x^2\ -\ 4x}{(x^2\ -\ 3)^2}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{-3\ -\ x^2\ -\ 4x}{(x^2\ -\ 3)^2}} \\ \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf \dfrac{dy}{dx}\ =\ \dfrac{-(3\ +\ x^2\ +\ 4x)}{(x^2\ -\ 3)^2}}}}}\bigstar$

$\bf{\underline{\underline{Now,\ finding\ the\ value\ of\ derivative\ at\ x\ =\ 0:-}}}$

$\bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{-(3\ +\ 0\ +\ 0)}{(0\ -\ 3)^2}} \\ \\ \\ \bf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{-3}{9}} \\ \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf \dfrac{dy}{dx}\ =\ \dfrac{-1}{3}}}}}\bigstar$