Question

refer to question and answer please..
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Answer 1

Given Expression,

$\sf y = \sqrt{ \cos(x) + \sqrt{ \cos(x) + \sqrt{ \cos(x) + \dots \infty } } }y= </p><p>cos(x)+ </p><p>cos(x)+ </p><p>cos(x)+…∞</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>The expression can be re written as :</p><p></p><p>[tex] \longrightarrow \sf \: y = \sqrt{ \cos(x) + y }⟶y= cos(x)+y</p><p> </p><p> </p><p></p><p>Squaring on both sides,</p><p></p><p>[tex] \longrightarrow \sf \: {y}^{2} = cos \: x + y⟶y </p><p>2</p><p> =cosx+y</p><p></p><p>[tex] \tt \pink{Differentiating w.r.t x,}$

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

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Formulas that can be Use:

• Derivative of cos x is - sin x
• Derivative of a constant is zero
• dxⁿ/dx = nxⁿ/x

Answer 2

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• Refer to the attachment plz..!!

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