The derivative of y = sqrt{sin x + sqrt{sin x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \infty}}}$.Since the expression repeats infinitely,we can write it as

Vedclass · Accepted Answer

$\frac{\cos x}{2y - 1}$

Vedclass · Answer

(A) Given the equation $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \infty}}}$.Since the expression repeats infinitely,we can write it as $y = \sqrt{\sin x + y}$.Squaring both sides,we get $y^2 = \sin x + y$.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$.This simplifies to $2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$.Rearranging the terms to solve for $\frac{dy}{dx}$,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$.Factoring out $\frac{dy}{dx}$,we have $\frac{dy}{dx}(2y - 1) = \cos x$.Therefore,$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$.

The derivative of $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \infty}}}$ is

Similar Questions

If $y = \sqrt{\log(x^2+1) + \sqrt{\log(x^2+1) + \sqrt{\log(x^2+1) + \dots \infty}}}$, $|x| < 1$, then $\frac{dy}{dx} = $

If $y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots \ldots \infty}}}$,then $\frac{dy}{dx}$ is equal to

If $y = \sqrt {x + \sqrt {x + \sqrt {x + \dots \infty } } }$,then $\frac{dy}{dx} = $

If $y = (\sqrt{x})^{(\sqrt{x})^{(\sqrt{x})^{\dots\infty}}}$,then $\frac{dy}{dx} = $

If $ y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} $,then $ \frac{d y}{d x}= $

Vedclass Test Series

Exam Paper Generator

Online Exam Module