Question

Solve the question in the attachment!​

Answer 1

Question :-

$\rm \:If \: y = {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + - - - \infty } } } } , \: then \: \dfrac{dy}{dx} =$

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

$\rm \: y = {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + - - - \infty } } } }$

can be rewritten as

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

can be further rewritten as

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

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

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}\bigg( {x}^{2} + {y}^{ - 1} \bigg)$

We know,

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

So, on using this result, we get

$\rm \: \dfrac{dy}{dx} = {2x}^{2 - 1} - {y}^{ - 1 - 1}\dfrac{dy}{dx}$

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

$\rm \: \dfrac{dy}{dx} = 2x - \dfrac{1}{ {y}^{2} } \dfrac{dy}{dx}$

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

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

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

$\rm\implies \:\dfrac{dy}{dx} = \dfrac{2x {y}^{2} }{ {y}^{2} + 1}$

[ Remark ]

Alter Solution

$\rm \: y = {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + \dfrac{1}{ {x}^{2} + - - - \infty } } } }$

can be rewritten as

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

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

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

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

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

$\rm \: 2y\dfrac{dy}{dx} = {x}^{2}\dfrac{dy}{dx} +2xy$

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

$\rm \: \bigg(2y - {x}^{2}\bigg)\dfrac{dy}{dx} = 2xy$

$\rm\implies \:\dfrac{dy}{dx} = \dfrac{2xy}{2y - {x}^{2} }$

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Additional Information

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

Answer 2

Step-by-step explanation:

$\color{purple} \begin{array}{ |c| } \hline \text{\boxed{\sf Solution.} Given} \tt\quad y=x^{2}+\dfrac{1}{x^{2}+\dfrac{1}{x^{2}+\ldots \infty}}\\ \\ \tt \Rightarrow \quad y=x^{2}+\frac{1}{y} \Rightarrow x^{2}=y-\frac{1}{y} \\ \\ \tt \Rightarrow \quad y^{2}=y x^{2}+1 \\ \\ \tt\text{differentiating w.r.t. x} \\ \\ \tt \Rightarrow \quad \dfrac{d y}{d x}\left(2 y-x^{2}\right)=y \cdot 2 x+x^{2} \cdot \dfrac{d y}{d x}+0 \\ \\ \tt\Rightarrow \quad \dfrac{d y}{d x}=\dfrac{2 x y}{2 y-x^{2}} \\ \\ \tt \Rightarrow \quad \dfrac{d y}{d x}=\cfrac{2 x y}{2 y-y+\dfrac{1}{y}} \\ \\ \tt \Rightarrow \quad \dfrac{d y}{d x}=\dfrac{2 x y^{2}}{1+y^{2}} \end{array}$