Question

find dy/dx if y =1/x²+3 class 11 standard​

Answer 1

Given Expression:-

$\\\quad\bullet\quad\sf y=\dfrac{1}{x^{2}+3 }$

and we are asked to find dy/dx

Solution:-

$\\\quad\longrightarrow\quad\sf y = \dfrac{1}{x^{2} +3}$

$\\\quad\longrightarrow\quad\sf y = \bigg(x^{2} +3\bigg)^{-1}$

Now differentiate the above equation with respect to x

$\\\quad\longrightarrow\quad\sf \dfrac{dy}{dx} = -1\bigg(x^{2} +3\bigg)^{-1-1} \bigg(2x^{2-1}+0 \bigg)$

$\\\quad\longrightarrow\quad\sf \dfrac{dy}{dx} = -1\bigg(x^{2} +3\bigg)^{-2} \bigg(2x\bigg)$

$\\\quad\longrightarrow\quad \boxed{\sf \dfrac{dy}{dx} = \dfrac{-2x}{(x^{2} +3)^{2} }}\bigstar$

Applied Concepts:-

• Chain Rule of differentiation

• dy/dx = n[a^(n-1)] for y = a^n

________________________________

Answer 2

Answer:

$\mathrm{\dfrac{-2x}{(x^2 + 3)^2}}$

Step-by-step explanation:

Given :

$\mathrm{y = \dfrac{1}{x^2+3}}$

To find :

$\mathrm{\dfrac{dy}{dx}}$

Solution :

$\longmapsto \mathrm{y = \dfrac{1}{x^2+3}}$

$\implies \mathrm{\dfrac{dy}{dx} = \dfrac{d}{dx}\bigg(\dfrac{1}{x^2+3}\bigg)}$

Let x² + 3 = u

So, By chain rule of differentiation,

$\implies \mathrm{\dfrac{dy}{dx} = \dfrac{d}{dx}\bigg(\dfrac{1}{u}\bigg)}$

We know that,

$\boxed{\mathrm{\dfrac{d}{dx}x^n = nx^{n-1}}}$

Here, n = -1

So, n-1 = -1-1 = -2

$\implies \mathrm{\dfrac{d}{dx}\bigg(\dfrac{1}{u}\bigg) = \dfrac{-1}{u^2}\dfrac{du}{dx}}$

Recalling, u = x² + 3; we get,

$\implies \mathrm{\dfrac{-1}{(x^2 + 3)^2}\times \dfrac{d(x^2+3)}{dx}}$

$\implies \mathrm{\dfrac{-1}{(x^2 + 3)^2}\times 2x}$

$\implies \mathrm{\dfrac{-2x}{(x^2 + 3)^2}}$

$\therefore \underline{\boxed{\mathbf{\dfrac{dy}{dx} = \dfrac{-2x}{(x^2+3)^2}}}}$

$\bigstar \textsf{ \underline{Some basic differentiations} :}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf x^n & \sf nx^{n-1} \\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x} \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx \end{array}} \\ \end{gathered}$

Hope it helps!!