To find $y''$ by implicit differentiation for the equation: $4x^2 + y^2 = 3$

we’ll proceed in the following steps:

Step 1: Differentiate Both Sides with Respect to $x$

Differentiating implicitly: $\frac{d}{dx}(4x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(3)$

1. The derivative of $4x^2$ with respect to $x$ is $8x$.
2. For $y^2$, use the chain rule: $\frac{d}{dx}(y^2) = 2y \cdot y'$.
3. The derivative of the constant $3$ is $0$.

This gives: $8x + 2y \cdot y' = 0$

Step 2: Solve for $y'$

Rearrange to isolate $y'$: $2y \cdot y' = -8x$ $y' = \frac{-8x}{2y} = \frac{-4x}{y}$

Step 3: Differentiate $y'$ to Find $y''$

Now, differentiate $y' = \frac{-4x}{y}$ with respect to $x$ again to find $y''$. Use the quotient rule, where if $u = -4x$ and $v = y$, then $y' = \frac{u}{v}$ and: $y'' = \frac{v \cdot u' - u \cdot v'}{v^2}$

1. $u = -4x$ so $u' = -4$.
2. $v = y$ so $v' = y' = \frac{-4x}{y}$.

Substitute these values: $y'' = \frac{y \cdot (-4) - (-4x) \cdot \frac{-4x}{y}}{y^2}$ Simplify each term: $y'' = \frac{-4y - \frac{16x^2}{y}}{y^2}$ Combine terms: $y'' = \frac{-4y^2 - 16x^2}{y^3}$

So, the second derivative $y''$ is: $y'' = \frac{-4(y^2 + 4x^2)}{y^3}$

Since $4x^2 + y^2 = 3$ from the original equation, substitute $y^2 + 4x^2 = 3$: $y'' = \frac{-4 \cdot 3}{y^3} = \frac{-12}{y^3}$

Final Answer

$y'' = \frac{-12}{y^3}$

Would you like further details or have any questions?

Here are some related questions for practice:

1. How does implicit differentiation differ from regular differentiation?
2. Why do we use the chain rule when differentiating $y^2$ with respect to $x$?
3. Can you apply the quotient rule to differentiate other types of equations?
4. What are the steps to confirm if $y$ is a function of $x$ in an implicit function?
5. How would this process change if the equation included $xy$ terms?

Tip: Always remember to use implicit differentiation when $y$ is a function of $x$ in terms that aren't easily separated.