y^4-2x=5, find d^2y/d^2x at (-2,1)

Implicit Differentiation: To find the second derivative of y with respect to x, we first need to implicitly differentiate the given equation with respect to x. The given equation is y^4 - 2x = 5. Differentiating both sides with respect to x, we get: 4y^3 * (dy/dx) - 2 = 0 Now, we solve for dy/dx: 4y^3 * (dy/dx) = 2 (dy/dx) = 2 / (4y^3) (dy/dx) = 1 / (2y^3)Solving for dy/dx: Next, we need to differentiate (dy/dx) with respect to x again to find the second derivative d^2y/dx^2. This requires using the chain rule since y is a function of x. Differentiating 1 / (2y^3) with respect to x, we get: d^2y/dx^2 = -3 * (1 / (2y^4)) * (dy/dx)Finding Second Derivative: We already found that (dy/dx) = 1 / (2y^3), so we substitute this into our expression for the second derivative: d^2y/dx^2 = -3 * (1 / (2y^4)) * (1 / (2y^3)) d^2y/dx^2 = -3 / (4y^7)Evaluating at (-2,1): Now we need to evaluate the second derivative at the point (-2,1). We substitute y = 1 into the expression for the second derivative: d^2y/dx^2 = -3 / (4 * 1^7) d^2y/dx^2 = -3 / 4

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$y^4-2x=5$ , find $\frac{d^2y}{d^2x}$ at $(-2,1)$

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y^4-2x=5

, find

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

(-2,1)

Implicit Differentiation: To find the second derivative of $y$ with respect to $x$ , we first need to implicitly differentiate the given equation with respect to $x$ . The given equation is $y^4 - 2x = 5$ . Differentiating both sides with respect to $x$ , we get: $4y^3 \cdot \left(\frac{dy}{dx}\right) - 2 = 0$ Now, we solve for $\frac{dy}{dx}$ : $4y^3 \cdot \left(\frac{dy}{dx}\right) = 2$ $\frac{dy}{dx} = \frac{2}{4y^3}$ $\frac{dy}{dx} = \frac{1}{2y^3}$
Solving for $\frac{dy}{dx}$ : Next, we need to differentiate $\frac{dy}{dx}$ with respect to $x$ again to find the second derivative $\frac{d^2y}{dx^2}$ . This requires using the chain rule since $y$ is a function of $x$ . $\newline$ Differentiating $\frac{1}{2y^3}$ with respect to $x$ , we get: $\newline$ $\frac{d^2y}{dx^2} = -3 \cdot \left(\frac{1}{2y^4}\right) \cdot \left(\frac{dy}{dx}\right)$
Finding Second Derivative: We already found that $(\frac{dy}{dx}) = \frac{1}{2y^3}$ , so we substitute this into our expression for the second derivative: $\newline$ $\frac{d^2y}{dx^2} = -3 \times \left(\frac{1}{2y^4}\right) \times \left(\frac{1}{2y^3}\right)$ $\newline$ $\frac{d^2y}{dx^2} = \frac{-3}{4y^7}$
Evaluating at $(-2,1)$ : Now we need to evaluate the second derivative at the point $(-2,1)$ . We substitute $y = 1$ into the expression for the second derivative: $\newline$ $\frac{d^2y}{dx^2} = -\frac{3}{4 \cdot 1^7}$ $\newline$ $\frac{d^2y}{dx^2} = -\frac{3}{4}$