Find y^('') if x^(4)+y^(4)=16

Differentiate Equation: Differentiate both sides of the equation with respect to x. Given: $x^4 + y^4 = 16$ Differentiate: $\frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = \frac{d}{dx}(16)$ Using the chain rule on $y^4$: $4x^3 + 4y^3 \frac{dy}{dx} = 0$Solve for dy/dx: Solve for $\frac{dy}{dx}$ (first derivative of y). Rearrange the differentiated equation: $4y^3 \frac{dy}{dx} = -4x^3$ $\frac{dy}{dx} = -\frac{x^3}{y^3}$Second Derivative of y: Differentiate $\frac{dy}{dx}$ again to find $\frac{d^2y}{dx^2}$ (second derivative of y). Differentiate: $\frac{d}{dx}\left(-\frac{x^3}{y^3}\right)$ Using the quotient rule: $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 - (-x^3)(3y^2)\frac{dy}{dx}}{y^6}$ Substitute $\frac{dy}{dx} = -\frac{x^3}{y^3}$ into the equation. $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 + 3x^3y^2 \left(-\frac{x^3}{y^3}\right)}{y^6}$ Simplify: $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 - 3x^6}{y^6}$

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find $y^{\prime \prime}$ if $x^{4}+y^{4}=16$

View step-by-step help

Home

Math Problems

Algebra 2

Composition of linear and quadratic functions: find a value

Full solution

Q. Find

y^{\prime \prime}

x^{4}+y^{4}=16

Differentiate Equation: Differentiate both sides of the equation with respect to x. $\newline$ Given: $x^4 + y^4 = 16$ $\newline$ Differentiate: $\frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = \frac{d}{dx}(16)$ $\newline$ Using the chain rule on $y^4$ : $4x^3 + 4y^3 \frac{dy}{dx} = 0$
Solve for dy/dx: Solve for $\frac{dy}{dx}$ (first derivative of y). $\newline$ Rearrange the differentiated equation: $4y^3 \frac{dy}{dx} = -4x^3$ $\newline$ $\frac{dy}{dx} = -\frac{x^3}{y^3}$
Second Derivative of y: Differentiate $\frac{dy}{dx}$ again to find $\frac{d^2y}{dx^2}$ (second derivative of y). $\newline$ Differentiate: $\frac{d}{dx}\left(-\frac{x^3}{y^3}\right)$ $\newline$ Using the quotient rule: $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 - (-x^3)(3y^2)\frac{dy}{dx}}{y^6}$ $\newline$ Substitute $\frac{dy}{dx} = -\frac{x^3}{y^3}$ into the equation. $\newline$ $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 + 3x^3y^2 \left(-\frac{x^3}{y^3}\right)}{y^6}$ $\newline$ Simplify: $\frac{d^2y}{dx^2} = \frac{-3x^2y^3 - 3x^6}{y^6}$