If y = frac{sqrt{a + x} - sqrt{a - x}}{sqrt{a | vedclass

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

Question

(A) Given $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$.Rationalizing the denominator,we multiply the numerator and denominato

Vedclass · Accepted Answer

$\frac{ay}{x\sqrt{a^2 - x^2}}$

Vedclass · Answer

(A) Given $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{a + x} - \sqrt{a - x})$:$y = \frac{(\sqrt{a + x} - \sqrt{a - x})^2}{(a + x) - (a - x)} = \frac{(a + x) + (a - x) - 2\sqrt{a^2 - x^2}}{2x} = \frac{2a - 2\sqrt{a^2 - x^2}}{2x} = \frac{a - \sqrt{a^2 - x^2}}{x} \dots (i)$Now,differentiating with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(a - \sqrt{a^2 - x^2}) - (a - \sqrt{a^2 - x^2}) \cdot \frac{d}{dx}(x)}{x^2}$$\frac{dy}{dx} = \frac{x \cdot [0 - \frac{1}{2\sqrt{a^2 - x^2}} \cdot (-2x)] - (a - \sqrt{a^2 - x^2})}{x^2}$$\frac{dy}{dx} = \frac{\frac{x^2}{\sqrt{a^2 - x^2}} - a + \sqrt{a^2 - x^2}}{x^2} = \frac{x^2 - a\sqrt{a^2 - x^2} + a^2 - x^2}{x^2\sqrt{a^2 - x^2}} = \frac{a^2 - a\sqrt{a^2 - x^2}}{x^2\sqrt{a^2 - x^2}}$$\frac{dy}{dx} = \frac{a(a - \sqrt{a^2 - x^2})}{x^2\sqrt{a^2 - x^2}} = \frac{a}{x\sqrt{a^2 - x^2}} \cdot \left( \frac{a - \sqrt{a^2 - x^2}}{x} \right)$Substituting from $(i)$,we get $\frac{dy}{dx} = \frac{ay}{x\sqrt{a^2 - x^2}}$.

If $y = \frac{\sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}}$,then $\frac{dy}{dx} = $

Similar Questions

If $y = \frac{a^{\cos^{-1}x}}{1 + a^{\cos^{-1}x}}$ and $z = a^{\cos^{-1}x}$,then $\frac{dy}{dz} = $

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \sec^{-1} \left( \frac{1 + x^2}{1 - x^2} \right)$,then $\frac{dy}{dx} =$

If $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$,then $f^{\prime}(x) = $

If $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$ and $x \in \left( {0,\frac{\pi }{2}} \right)$,then $\frac{{dy}}{{dx}}$ is equal to

For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}$, the value of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module