Question

important question for 12th board .

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Answer 1

AnswEr :

Given Expression,

$\sf y = \sqrt{ \cos(x) + \sqrt{ \cos(x) + \sqrt{ \cos(x) + \dots \infty } } }$

The expression can be re written as :

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

Squaring on both sides,

$\longrightarrow \sf \: {y}^{2} = cos \: x + y$

Differentiating w.r.t x,

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

Formulas Used

• Derivative of cos x is - sin x

• Derivative of a constant is zero

• dxⁿ/dx = nxⁿ/x

Answer 2

Given :

If $\sf\:y=\sqrt{\cos\:x+\sqrt{\cos\:x+\sqrt{\cos\:x+...\infty}}}$

To Find :

dy/dx

Formula :

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

• Differentiation Formula's

$1) \dfrac{d(sinx)}{dx} = cosx$

$2) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

$3) \dfrac{d(constant)}{dx} = 0$

Solution :

$\sf\:y=\sqrt{\cos\:x+\sqrt{\cos\:x+\sqrt{\cos\:x+...\infty}}}$

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

Now squaring on both sides

$\sf\implies\:y^2=\cos\:x+y$

$\sf\implies\:y^2-y=x$

Now differentiate with respect to x

$\sf\implies2y\dfrac{dy}{dx}-\dfrac{dy}{dx}=-sin\:x$

$\sf\implies(2y-1)\dfrac{dy}{dx}=-\sin\:x$

$\sf\implies\dfrac{dy}{dx}=\dfrac{-\sin\:x}{2y-1}$

$\sf\implies\dfrac{dy}{dx}=\dfrac{\sin\:x}{1-2y}$

Hence proved !