Let x^(2)-y^(2)=28. What is the value of (d^(2)y)/(dx^(2)) at the point (8,-6) ? Give an exact number.

Differentiate Once: We are given the equation x^(2) - y^(2) = 28. To find (d^(2)y)/(dx^(2)), we need to differentiate the equation with respect to x twice. First, let's differentiate the equation once with respect to x. Using the chain rule, the derivative of x^(2) with respect to x is 2x, and the derivative of y^(2) with respect to x is 2y(dy/dx). So, the first derivative of the equation is: 2x - 2y(dy/dx) = 0Solve for dy/dx: Now, we need to solve for dy/dx. Rearrange the equation to isolate dy/dx: 2y(dy/dx) = 2x dy/dx = 2x / (2y) dy/dx = x / yDifferentiate Again: Next, we differentiate the equation again with respect to x to find the second derivative, d^(2)y/dx^(2). Differentiating x/y with respect to x using the quotient rule, which is (v(du/dx) - u(dv/dx)) / v^2, where u = x and v = y, we get: d^(2)y/dx^(2) = (y(1) - x(dy/dx)) / y^2Substitute dy/dx: We already found that dy/dx = x / y, so we substitute this into our expression for the second derivative: d^(2)y/dx^(2) = (y - x(x/y)) / y^2 d^(2)y/dx^(2) = (y - x^2/y) / y^2Simplify Expression: Now we simplify the expression: d^(2)y/dx^(2) = (y^2 - x^2) / y^3 Since we know from the original equation that x^2 - y^2 = 28, we can rewrite y^2 - x^2 as -28. d^(2)y/dx^(2) = (-28) / y^3Substitute Point: Finally, we substitute the given point (8, -6) into the expression to find the exact value of the second derivative at that point. d^(2)y/dx^(2) = (-28) / (-6)^3 d^(2)y/dx^(2) = (-28) / (-216) d^(2)y/dx^(2) = 28 / 216Final Answer: Simplify the fraction to get the final answer: d^(2)y/dx^(2) = 1 / 7.7

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Let
x^(2)-y^(2)=28.
What is the value of
(d^(2)y)/(dx^(2)) at the point
(8,-6) ?
Give an exact number.

Let $x^{2}-y^{2}=28$ . $\newline$ What is the value of $\frac{d^{2} y}{d x^{2}}$ at the point $(8,-6)$ ? $\newline$ Give an exact number.

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Q. Let

x^{2}-y^{2}=28

\newline

What is the value of

\frac{d^{2} y}{d x^{2}}

at the point

(8,-6)

\newline

Give an exact number.

Differentiate Once: We are given the equation $x^{2} - y^{2} = 28$ . To find $\frac{d^{2}y}{dx^{2}}$ , we need to differentiate the equation with respect to $x$ twice. $\newline$ First, let's differentiate the equation once with respect to $x$ . $\newline$ Using the chain rule, the derivative of $x^{2}$ with respect to $x$ is $2x$ , and the derivative of $y^{2}$ with respect to $x$ is $2y\frac{dy}{dx}$ . $\newline$ So, the first derivative of the equation is: $\newline$ $\frac{d^{2}y}{dx^{2}}$ $0$
Solve for $\frac{dy}{dx}$ : Now, we need to solve for $\frac{dy}{dx}$ . Rearrange the equation to isolate $\frac{dy}{dx}$ : $2y\left(\frac{dy}{dx}\right) = 2x$ $\frac{dy}{dx} = \frac{2x}{2y}$ $\frac{dy}{dx} = \frac{x}{y}$
Differentiate Again: Next, we differentiate the equation again with respect to $x$ to find the second derivative, $\frac{d^2y}{dx^2}$ . Differentiating $\frac{x}{y}$ with respect to $x$ using the quotient rule, which is $\frac{v\left(\frac{du}{dx}\right) - u\left(\frac{dv}{dx}\right)}{v^2}$ , where $u = x$ and $v = y$ , we get: $\frac{d^2y}{dx^2} = \frac{y(1) - x\left(\frac{dy}{dx}\right)}{y^2}$
Substitute $\frac{dy}{dx}$ : We already found that $\frac{dy}{dx} = \frac{x}{y}$ , so we substitute this into our expression for the second derivative: $\newline$ $\frac{d^2y}{dx^2} = \frac{y - x\left(\frac{x}{y}\right)}{y^2}$ $\newline$ $\frac{d^2y}{dx^2} = \frac{y - \frac{x^2}{y}}{y^2}$
Simplify Expression: Now we simplify the expression: $\newline$ $\frac{d^2y}{dx^2} = \frac{y^2 - x^2}{y^3}$ $\newline$ Since we know from the original equation that $x^2 - y^2 = 28$ , we can rewrite $y^2 - x^2$ as $-28$ . $\newline$ $\frac{d^2y}{dx^2} = \frac{-28}{y^3}$
Substitute Point: Finally, we substitute the given point $(8, -6)$ into the expression to find the exact value of the second derivative at that point. $\newline$ $\frac{d^2y}{dx^2} = \frac{-28}{(-6)^3}$ $\newline$ $\frac{d^2y}{dx^2} = \frac{-28}{-216}$ $\newline$ $\frac{d^2y}{dx^2} = \frac{28}{216}$
Final Answer: Simplify the fraction to get the final answer: $\frac{d^{2}y}{dx^{2}} = \frac{1}{7.7}$