If x = ysqrt{1 - y^2},then frac{dy}{dx} = | vedclass

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

Question

(C) Given equation: $x = y\sqrt{1 - y^2}$Differentiating both sides with respect to $x$ using the product rule and chain rule:$1 = \frac{dy}{dx} \cdot

Vedclass · Accepted Answer

$\frac{\sqrt{1 - y^2}}{1 - 2y^2}$

Vedclass · Answer

(C) Given equation: $x = y\sqrt{1 - y^2}$Differentiating both sides with respect to $x$ using the product rule and chain rule:$1 = \frac{dy}{dx} \cdot \sqrt{1 - y^2} + y \cdot \frac{1}{2\sqrt{1 - y^2}} \cdot (-2y) \cdot \frac{dy}{dx}$Simplify the expression:$1 = \frac{dy}{dx} \left( \sqrt{1 - y^2} - \frac{y^2}{\sqrt{1 - y^2}} \right)$Find a common denominator inside the bracket:$1 = \frac{dy}{dx} \left( \frac{1 - y^2 - y^2}{\sqrt{1 - y^2}} \right)$$1 = \frac{dy}{dx} \left( \frac{1 - 2y^2}{\sqrt{1 - y^2}} \right)$Solving for $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{1 - 2y^2}$

If $x = y\sqrt{1 - y^2}$,then $\frac{dy}{dx} = $

Similar Questions

Let $f(x), x \in [0, \infty)$ be a non-negative continuous function. If $f'(x) \cos x \le f(x) \sin x$ for all $x \ge 0$,then the value of $f(2\pi)$ is equal to

If $\tan (x + y) + \tan (x - y) = 1,$ then $\frac{dy}{dx} = $

If $x = \exp \left\{ {{{\tan }^{ - 1}}\left( {{{y - {x^2}} \over {{x^2}}}} \right)} \right\}$,then $\frac{dy}{dx}$ equals

If $ \cos y = x \cos (a+y) $ with $ \cos a \neq \pm 1 $,then $ \frac{d y}{d x} $ is equal to

If $3 \sin xy + 4 \cos xy = 5$,then $\frac{dy}{dx}$ is equal to . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module