Math, asked by mansi5228, 5 hours ago

refer to question and answer please..kindly don't be greedy of thanks.. Urgent work needed... ​

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Answers

Answered by senboni123456
3

Step-by-step explanation:

We have,

y =  \sqrt{ \cos(x) +  \sqrt{ \cos(x)  +  \sqrt{ \cos(x) + ... }  }  }  \\

 \implies \: y =  \sqrt{ \cos(x) +y }  \\

 \implies \:  y ^{2} =   \cos(x) +y  \\

 \implies \: y ^{2}  - y -  \cos(x)  = 0  \\

 \implies \: y  =  \frac{1 \pm \sqrt{1 + 4 \cos(x) } }{2}  \\

Answered by TrustedAnswerer19
43

\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}

Given,

 \sf y = \sqrt{ \cos(x) + \sqrt{ \cos(x) + \sqrt{ \cos(x) + \dots \infty } } }</p><p></p><p></p><p>[tex] \longrightarrow \sf \: y = \sqrt{ \cos(x) + y }⟶y= cos(x)+y</p><p></p><p>[tex] Squaring on both sides,

 \longrightarrow \sf \: {y}^{2} = cos \: x + y⟶y 2 =cosx+y</p><p></p><p>[tex] Differentiating w.r.t x,

 \begin{gathered} \longrightarrow \sf \: \dfrac{ {dy}^{2} }{dx} = \dfrac{d(cos \: x)}{dx} + \dfrac{dy}{dx} \\ \\ \longrightarrow \sf \: 2y \dfrac{dy}{dx} = - sin \: x + \dfrac{dy}{dx} \\ \\ \longrightarrow \sf 2y \dfrac{dy}{dx} - \dfrac{dy}{dx} = - sin \ x\\ \\ \longrightarrow \sf \: \dfrac{dy}{dx} (2y - 1) = - sin \: x \\ \\ \longrightarrow \boxed{ \boxed{ \sf \frac{dy}{dx} = \dfrac{ - sin \: x}{2y - 1} }}\end{gathered}

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