If sqrt{x+y}+sqrt{y-x}=5,then left(frac{d^{2} | vedclass

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

Question

(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$. Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$. Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x

Vedclass · Accepted Answer

$\frac{2}{25}$

Vedclass · Answer

(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$. Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$. Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x+y}$. Simplifying,$-2x = 25 - 10\sqrt{x+y}$,which gives $10\sqrt{x+y} = 2x + 25$. Differentiating both sides with respect to $x$: $10 	imes \frac{1}{2\sqrt{x+y}} 	imes (1 + \frac{dy}{dx}) = 2$. $\frac{5}{\sqrt{x+y}} 	imes (1 + \frac{dy}{dx}) = 2$. $1 + \frac{dy}{dx} = \frac{2\sqrt{x+y}}{5}$. Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{2}{5} 	imes \frac{1}{2\sqrt{x+y}} 	imes (1 + \frac{dy}{dx})$. Substituting the value of $(1 + \frac{dy}{dx})$ from the previous step: $\frac{d^2y}{dx^2} = \frac{1}{5\sqrt{x+y}} 	imes \frac{2\sqrt{x+y}}{5} = \frac{2}{25}$.

If $\sqrt{x+y}+\sqrt{y-x}=5$,then $\left(\frac{d^{2} y}{d x^{2}}\right)=$

Similar Questions

$y=e^{a \sin ^{-1} x} \Rightarrow (1-x^2) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to

If $y = \sin(m \sin^{-1} x)$, then $(1-x^2) y_2 - x y_1$ is equal to (Here, $y_n$ denotes $\frac{d^n y}{dx^n}$)

For $y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\} ;|x| \leq 1$, if $a\left(1-x^{2}\right) y_{2}+b x y_{1}=0$ then $(a, b)=$

If $y=3 \cos (\log x)+4 \sin (\log x)$,then $x^2 y_2+x y_1=$

If $x = \log t, t > 0$ and $y = \frac{1}{t}$,then $\frac{d^2 y}{d x^2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module