Question

if y=√x^2+1.differentiate using chain rule.
I'm gud. gud aftr noon. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: y = \sqrt{ {x}^{2} + 1}$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} }$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} }$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} - 1}\dfrac{d}{dx}\bigg[ {x}^{2} + 1\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}(f + g) = \dfrac{d}{dx}f + \dfrac{d}{dx}g}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ \dfrac{d}{dx}{x}^{2} + \dfrac{d}{dx}1\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ 2x + 0\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ 2x\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ x\bigg]$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{x}{ \sqrt{ {x}^{2} + 1} }}}$

More to know :-

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

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: y = \sqrt{ {x}^{2} + 1}$

can be rewritten as

$\rm :\longmapsto\:y = {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} }$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} }$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{\dfrac{1}{2} - 1}\dfrac{d}{dx}\bigg[ {x}^{2} + 1\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}(f + g) = \dfrac{d}{dx}f + \dfrac{d}{dx}g}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ \dfrac{d}{dx}{x}^{2} + \dfrac{d}{dx}1\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ 2x + 0\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ 2x\bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {\bigg[ {x}^{2} + 1 \bigg]}^{ - \: \dfrac{1}{2}}\bigg[ x\bigg]$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{x}{ \sqrt{ {x}^{2} + 1} }}}$

More to know :-

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