important question for 12th board .
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AnswEr :
Given Expression,
[tex] \sf y = \sqrt{ \cos(x) + \sqrt{ \cos(x) + \sqrt{ \cos(x) + \dots \infty } } } [/tex]
The expression can be re written as :
[tex] \longrightarrow \sf \: y = \sqrt{ \cos(x) + y } [/tex]
Squaring on both sides,
[tex] \longrightarrow \sf \: {y}^{2} = cos \: x + y [/tex]
Differentiating w.r.t x,
[tex] \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} }}[/tex]
Formulas Used
Given :
If [tex]\sf\:y=\sqrt{\cos\:x+\sqrt{\cos\:x+\sqrt{\cos\:x+...\infty}}}[/tex]
To Find :
dy/dx
Formula :
• Chain rule
Let y=f(t) ,t = g(u) and u =m(x) ,then
[tex] \dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx} [/tex]
• Differentiation Formula's
[tex]1) \dfrac{d(sinx)}{dx} = cosx[/tex]
[tex]2) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1} [/tex]
[tex]3) \dfrac{d(constant)}{dx} = 0[/tex]
Solution :
[tex]\sf\:y=\sqrt{\cos\:x+\sqrt{\cos\:x+\sqrt{\cos\:x+...\infty}}}[/tex]
[tex]\sf\implies\:y=\sqrt{\cos\:x+y}[/tex]
Now squaring on both sides
[tex]\sf\implies\:y^2=\cos\:x+y[/tex]
[tex]\sf\implies\:y^2-y=x[/tex]
Now differentiate with respect to x
[tex]\sf\implies2y\dfrac{dy}{dx}-\dfrac{dy}{dx}=-sin\:x[/tex]
[tex]\sf\implies(2y-1)\dfrac{dy}{dx}=-\sin\:x[/tex]
[tex]\sf\implies\dfrac{dy}{dx}=\dfrac{-\sin\:x}{2y-1}[/tex]
[tex]\sf\implies\dfrac{dy}{dx}=\dfrac{\sin\:x}{1-2y}[/tex]
Hence proved !