Question

If $\sqrt{x+y}+\sqrt{y-x}=5$, then $25\frac{d^2y}{dx^2}$ is equal to

Answer 1

2

Answer 2

To find $25\frac{d^2y}{dx^2}$, we first need to find an explicit expression for $y$ in terms of $x$ from the given equation.

Given equation: $\sqrt{x+y}+\sqrt{y-x}=5$

Step 1: Simplify the equation to express $y$ in terms of $x$. Square both sides of the equation: $(\sqrt{x+y}+\sqrt{y-x})^2 = 5^2$ Using the identity $(a+b)^2 = a^2+b^2+2ab$: $(x+y) + (y-x) + 2\sqrt{(x+y)(y-x)} = 25$ $2y + 2\sqrt{y^2-x^2} = 25$

Isolate the square root term: $2\sqrt{y^2-x^2} = 25 - 2y$

Square both sides again: $(2\sqrt{y^2-x^2})^2 = (25 - 2y)^2$ $4(y^2-x^2) = 625 - 100y + 4y^2$ $4y^2 - 4x^2 = 625 - 100y + 4y^2$

Cancel $4y^2$ from both sides: $-4x^2 = 625 - 100y$

Rearrange the terms to solve for $y$: $100y = 625 + 4x^2$ $y = \frac{625 + 4x^2}{100}$ $y = \frac{625}{100} + \frac{4x^2}{100}$ $y = \frac{25}{4} + \frac{x^2}{25}$

Step 2: Find the first derivative $\frac{dy}{dx}$. Differentiate $y = \frac{25}{4} + \frac{x^2}{25}$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}\left(\frac{25}{4}\right) + \frac{d}{dx}\left(\frac{x^2}{25}\right)$ $\frac{dy}{dx} = 0 + \frac{1}{25}(2x)$ $\frac{dy}{dx} = \frac{2x}{25}$

Step 3: Find the second derivative $\frac{d^2y}{dx^2}$. Differentiate $\frac{dy}{dx} = \frac{2x}{25}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2x}{25}\right)$ $\frac{d^2y}{dx^2} = \frac{2}{25}$

Step 4: Calculate $25\frac{d^2y}{dx^2}$. $25\frac{d^2y}{dx^2} = 25 \times \frac{2}{25}$ $25\frac{d^2y}{dx^2} = 2$