Question

If y = \[\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+....\infty }}}\] , then value of ( 2y – 1 ) \[\dfrac{dy}{dx}\] is equal to ?
(a) sin x
(b) cos x
(c) -sin x
(d) -cos x

Answer 1

Hint: This is a question of continuity and differentiability. The concept of differentiation deals with the idea of finding the derivative of a function. It is the process of determining the rate of change in functions on the basis of its variables. To solve this question, first, we square both sides of the question. Then we differentiate the equation and solve the equation then we find out the value of ( 2y – 1 ) $\dfrac{dy}{dx}$.

Formula Used: $\dfrac{d(\sin x)}{dx}=\cos x$

Complete step by step Solution:
We have given the question y = \[\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+....\infty }}}\]
We have to find the value of ( 2y – 1 ) $\dfrac{dy}{dx}$
Let y =\[\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+....\infty }}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Squaring both sides of equation (1), we get
${{y}^{2}}=\sin x+y$
Now differentiate both sides of the above equation w.r.t x, we get
$\dfrac{d}{dx}({{y}^{2}})=\dfrac{d}{dx}(\sin x+y)$
Solving both sides of the equation, we get
$2y.\dfrac{dy}{dx}=\cos x+\dfrac{dy}{dx}$
Now subtracting $\dfrac{dy}{dx}$from both sides , we get
$2y\dfrac{dy}{dx}-\dfrac{dy}{dx}=\cos x$
Solving the above equation, we get
( 2y – 1) $\dfrac{dy}{dx}$ = cos x
Hence ( 2y - 1 ) $\dfrac{dy}{dx}$ = cos x

Hence, the correct option is (b).

Note: Students make mistake in differentiating the equation . When we differentiate the equation , always remind to subtract 1 from the power . By doing a lot of practice, student will be able to solve the differentiation easily and accurately .