Question

in differention chapter

Answer 1

Question :

Find $\dfrac{dy}{dx}$ ,if

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

Formula's :

• Some standard Formula's

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

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

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

•${\red{\boxed{\large{\bold{Chain\:Rule}}}}}$

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

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

Solution :

We to Find $\dfrac{dy}{dx}$, if

$y = \sqrt{sinx + \sqrt{sinx + .... + \infty } }$

$\implies y = \sqrt{sinx + y}$

Now squaring on both sides

$\implies \: y {}^{2} = sinx + y$

$\implies \: y {}^{2} - y = sinx$

Now differinate with respect to x

$\implies \: 2y \dfrac{dy}{dx} - \dfrac{dy}{dx} = cosx$

$\implies \dfrac{dy}{dx} (2y - 1) = cosx$

$\implies \: \dfrac{dy}{dx} = \dfrac{cosx}{2y - 1}$