QuestionFind $y^{\prime \prime}$ by implicit differentiation. $$2 x^{3}+5 y^{3}=4$$

Studdy Solution
Differentiate $y' = \frac{-2x^2}{5y^2}$ implicitly with respect to $x$ to find $y''$.
Using the quotient rule, where $u = -2x^2$ and $v = 5y^2$, we have: $$y'' = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$$
Calculate $\frac{du}{dx}$: $$\frac{du}{dx} = \frac{d}{dx}(-2x^2) = -4x$$
Calculate $\frac{dv}{dx}$ using the chain rule: $$\frac{dv}{dx} = \frac{d}{dx}(5y^2) = 10y \cdot \frac{dy}{dx} = 10y \cdot y'$$
Substitute $y' = \frac{-2x^2}{5y^2}$ into $\frac{dv}{dx}$: $$\frac{dv}{dx} = 10y \cdot \frac{-2x^2}{5y^2} = \frac{-20x^2y}{5y^2} = \frac{-4x^2}{y}$$
Substitute into the quotient rule: $$y'' = \frac{5y^2 \cdot (-4x) - (-2x^2) \cdot \left(\frac{-4x^2}{y}\right)}{(5y^2)^2}$$
Simplify: $$y'' = \frac{-20xy^2 + \frac{8x^4}{y}}{25y^4}$$
Further simplify: $$y'' = \frac{-20xy^3 + 8x^4}{25y^5}$$
The second derivative $y''$ is: $$y'' = \frac{-20xy^3 + 8x^4}{25y^5}$$