To find $y''$ by implicit differentiation for the equation:

$5x^2 + y^2 = 3$

we'll follow these steps:

Step 1: Differentiate both sides with respect to $x$.

Since we have an equation involving $x$ and $y$, we need to apply implicit differentiation:

$\frac{d}{dx}(5x^2 + y^2) = \frac{d}{dx}(3)$

Differentiating each term:

1. For $5x^2$: $\frac{d}{dx}(5x^2) = 10x$
2. For $y^2$: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} = 2y \cdot y'$ (using the chain rule)
3. For the constant 3: $\frac{d}{dx}(3) = 0$

Putting it all together:

$10x + 2y \cdot y' = 0$

Step 2: Solve for $y'$.

Rearrange the equation to isolate $y'$:

$2y \cdot y' = -10x$

$y' = \frac{-10x}{2y} = \frac{-5x}{y}$

Step 3: Differentiate $y'$ again to find $y''$.

Now, we differentiate $y' = \frac{-5x}{y}$ with respect to $x$ to get $y''$:

$y'' = \frac{d}{dx}\left(\frac{-5x}{y}\right)$

Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$, where $u = -5x$ and $v = y$:

1. $u' = -5$
2. $v' = y' = \frac{-5x}{y}$

Substitute into the quotient rule:

$y'' = \frac{(-5)y - (-5x)\left(\frac{-5x}{y}\right)}{y^2}$

Simplify each term in the numerator:

$y'' = \frac{-5y + \frac{25x^2}{y}}{y^2}$

Combine terms over a common denominator $y$:

$y'' = \frac{-5y^2 + 25x^2}{y^3}$

Final Answer:

$y'' = \frac{25x^2 - 5y^2}{y^3}$

Let me know if you'd like more details, or if you have questions!

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Here are some related questions:

1. How do you apply implicit differentiation to find higher-order derivatives?
2. How does the chain rule help in implicit differentiation?
3. What are other methods to solve for second derivatives of implicit functions?
4. Why is it necessary to use the quotient rule when differentiating $y' = \frac{-5x}{y}$?
5. Can you verify the solution by differentiating $y' = \frac{-5x}{y}$ step-by-step?

Tip: Remember, implicit differentiation treats $y$ as a function of $x$, allowing derivatives with respect to $x$ on both sides of an equation.