If sqrt{x+y}-sqrt{x-y}=c, then frac{d^2 y}{d | vedclass

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

Question

(B) Given that $\sqrt{x+y}-\sqrt{x-y}=c$. Squaring both sides, we get $(x+y)+(x-y)-2 \sqrt{x^2-y^2}=c^2$. $2x - c^2 = 2 \sqrt{x^2-y^2}$. Squaring both

Vedclass · Accepted Answer

$\frac{-c^4}{4 y^3}$

Vedclass · Answer

(B) Given that $\sqrt{x+y}-\sqrt{x-y}=c$. Squaring both sides, we get $(x+y)+(x-y)-2 \sqrt{x^2-y^2}=c^2$. $2x - c^2 = 2 \sqrt{x^2-y^2}$. Squaring both sides again, we get $4x^2 + c^4 - 4xc^2 = 4(x^2-y^2) = 4x^2 - 4y^2$. $c^4 - 4xc^2 = -4y^2$, which implies $4y^2 = 4xc^2 - c^4$. Differentiating with respect to $x$, we get $8y \frac{dy}{dx} = 4c^2$, so $2y \frac{dy}{dx} = c^2$. Differentiating again with respect to $x$, we get $2(\frac{dy}{dx})^2 + 2y \frac{d^2y}{dx^2} = 0$. Thus, $\frac{d^2y}{dx^2} = -\frac{(\frac{dy}{dx})^2}{y}$. Since $\frac{dy}{dx} = \frac{c^2}{2y}$, we have $\frac{d^2y}{dx^2} = -\frac{(c^2/2y)^2}{y} = -\frac{c^4}{4y^3}$. Therefore, option $B$ is correct.

If $\sqrt{x+y}-\sqrt{x-y}=c$, then $\frac{d^2 y}{d x^2}=$

Similar Questions

If ${x^p}{y^q} = {(x + y)^{p + q}}$,then $\frac{{{d^2}y}}{{d{x^2}}} = $

Let $f$ be a twice differentiable function such that $f^{\prime \prime}(x) = -f(x)$,$f^{\prime}(x) = g(x)$ and $h(x) = [f(x)]^2 + [g(x)]^2$. If $h(5) = 1$,then $h(10)$ is $\qquad$

If $y=(\tan^{-1} x)^{2}$,show that $(x^{2}+1)^{2} y_{2}+2 x(x^{2}+1) y_{1}=2$.

If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to

If $y^2 = a x^2 + 2 x + c$,then $y^3 \frac{d^2 y}{d x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module