If y = sqrt{(1 - x)(1 + x)},then | vedclass

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

Question

(B) Given $y = \sqrt{(1 - x)(1 + x)}$.Squaring both sides,we get $y^2 = (1 - x)(1 + x) = 1 - x^2$.Differentiating both sides with respect to $x$,we ge

Vedclass · Accepted Answer

$(1 - x^2)\frac{dy}{dx} + xy = 0$

Vedclass · Answer

(B) Given $y = \sqrt{(1 - x)(1 + x)}$.Squaring both sides,we get $y^2 = (1 - x)(1 + x) = 1 - x^2$.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(1 - x^2)$.Using the chain rule,$2y \frac{dy}{dx} = -2x$.Dividing by $2$,we get $y \frac{dy}{dx} = -x$.Multiply both sides by $y$,we get $y^2 \frac{dy}{dx} = -xy$.Since $y^2 = 1 - x^2$,substituting this gives $(1 - x^2) \frac{dy}{dx} = -xy$.Rearranging the terms,we get $(1 - x^2) \frac{dy}{dx} + xy = 0$.

If $y = \sqrt{(1 - x)(1 + x)}$,then

Similar Questions

If $x^{p}+y^{q}=(x+y)^{p+q}$,then $\frac{dy}{dx}$ is

The focal length of a mirror is given by $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$. In finding the values of $u$ and $v$, the errors are equal to $p$. Then, the relative error in $f$ is

If $3 f(x)-2 f\left(\frac{1}{x}\right)=x$,then $f^{\prime}(2)$ is

If $e^y + xy = e$, then the ordered pair $(\frac{dy}{dx}, \frac{d^2y}{dx^2})$ at $x = 0$ is equal to

The slope of the tangent to the curve $f(x) = \tanh^{-1}(\sin x)$ at $x = \pi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module