Question

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

Answer 1

Step-by-step explanation:

We have,

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

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

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

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

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

Answer 2

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

Given,

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

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

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