Question

If

$y = {(x + \sqrt{ {x}^{2} + 1 }) }^{m} \:$

prove that

$\frac{dy}{dx} = \frac{my}{ \sqrt{ {x}^{2} + 1} }$
​

Answer 1

Answer:

the answer is in the picture.

Step-by-step explanation:

if you find my answer any helpful then please do rank me as the brainliest.

Answer 2

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

Given expression is

$\rm \: y=  \: {(x + \sqrt{ {x}^{2} + 1} )}^{m}$

On taking log on both sides, we get

$\rm \: logy=  \: log {(x + \sqrt{ {x}^{2} + 1} )}^{m}$

We know,

$\boxed{\sf{  \:\rm \: log {x}^{y} = y \: logx \: }}$

So, using this result, we get

$\rm \: logy=  \:m \: log {(x + \sqrt{ {x}^{2} + 1} )}$

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

$\rm \: \dfrac{d}{dx}logy=  \:m \: \dfrac{d}{dx} \: log {(x + \sqrt{ {x}^{2} + 1} )}$

We know,

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx}logx \: = \: \frac{1}{x} \: \: }}$

So, using this result, we get

$\rm \: \dfrac{1}{y}\dfrac{dy}{dx} =m \: \dfrac{1}{x + \sqrt{ {x}^{2} + 1} } \dfrac{d}{dx}(x + \sqrt{ {x}^{2} + 1 } )$

We know,

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

and

$\boxed{\sf{  \:\rm \: \dfrac{d}{dx} \sqrt{x} \: = \: \frac{1}{2 \sqrt{x} } \: \: }}$

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = \: \dfrac{my}{x + \sqrt{ {x}^{2} + 1} } \bigg[1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1} } \: \dfrac{d}{dx}( {x}^{2} + 1) \bigg]$

$\rm \: \dfrac{dy}{dx} = \: \dfrac{my}{x + \sqrt{ {x}^{2} + 1} } \bigg[1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1} } \: (2x + 0) \bigg]$

$\rm \: \dfrac{dy}{dx} = \: \dfrac{my}{x + \sqrt{ {x}^{2} + 1} } \bigg[1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1} } \: (2x) \bigg]$

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

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

$\rm\implies \: \: \boxed{\sf{  \: \: \rm \: \dfrac{dy}{dx} = \: \dfrac{my}{ \sqrt{ {x}^{2} + 1} } \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

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