Math, asked by madhav5245, 2 days ago

Differentiate the following function with respect to x

y = x +  \frac{1}{x +  \frac{1}{x +  \frac{1}{x} +  -  -  -  \infty  }}

Answers

Answered by mathdude500
36

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

Given function is

\rm \: y = x + \dfrac{1}{x + \dfrac{1}{x + \dfrac{1}{x +  \dfrac{1}{x}  +  -  -  -  \infty } } }  \\

can be rewritten as

\rm \: y = x + \dfrac{1}{y}

\rm \: y = \dfrac{xy + 1}{y}

\rm \:  {y}^{2}  = xy + 1

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

\rm \:  \dfrac{d}{dx} {y}^{2}  = \dfrac{d}{dx}(xy + 1)

We know,

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

So, using this result, we get

\rm \: 2y\dfrac{dy}{dx} = \dfrac{d}{dx}(xy) + \dfrac{d}{dx}1

We know,

\boxed{\tt{  \: \dfrac{d}{dx}uv \:  =  \: u\dfrac{d}{dx}v \:  +  \: v\dfrac{d}{dx}u \: }} \\

and

\boxed{\tt{  \: \dfrac{d}{dx}k \:  =  \: 0 \: }} \\

So, using this result, we get

\rm \: 2y\dfrac{dy}{dx} = x\dfrac{d}{dx}y + y\dfrac{d}{dx}x + 0

\rm \: 2y\dfrac{dy}{dx} = x\dfrac{dy}{dx}+ y \times 1

\rm \: 2y\dfrac{dy}{dx}  -  x\dfrac{dy}{dx} =  y

\rm \: \dfrac{dy}{dx}(2y-  x )=  y

\bf\implies \:\dfrac{dy}{dx} =  \dfrac{y}{2y - x}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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}

Answered by jaswasri2006
26

\underline{ \pink{ \rm GIVEN \:  \: DATA\: \: : }}

Function is

 \large\bf y =  x + \frac{1}{x +  \frac{1}{x +  \frac{1}{x}  \: . \: . \: . \:  \infty } }

\underline{ \purple{ \rm TO \: \: FIND \: \: : }}

Differentiating the Function with respect to x

\underline{ \red{ \mathfrak{ FINAL  \:  \: ANSWER\: \: : }}}

 \rm  \frac{dy}{dx}  =  \frac{y}{2y - x}

 \\

For Clear Explanation, refer the given attachment

Attachments:
Similar questions