We are tasked with finding the second derivative $y''$ by implicit differentiation for the equation:

$7x^2 + y^2 = 5$

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

$\frac{d}{dx}(7x^2 + y^2) = \frac{d}{dx}(5)$

The right-hand side is constant, so its derivative is 0. Now differentiate each term on the left-hand side:

• For $7x^2$, the derivative is $14x$.
• For $y^2$, apply the chain rule: $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$, which gives $2y y'$.

So the first derivative equation becomes:

$14x + 2y \frac{dy}{dx} = 0$

Step 2: Solve for $y'$

$2y \frac{dy}{dx} = -14x$

$\frac{dy}{dx} = \frac{y'}{2y} = -\frac{7x}{y}$

Thus, the first derivative $y'$ is:

$y' = -\frac{7x}{y}$

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

Now we differentiate $y' = -\frac{7x}{y}$ implicitly to find $y''$.

$\frac{d}{dx} \left( -\frac{7x}{y} \right)$

Apply the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$, where $u = -7x$ and $v = y$.

• The derivative of $u = -7x$ is $u' = -7$.
• The derivative of $v = y$ is $v' = y' = -\frac{7x}{y}$.

Substituting these into the quotient rule:

$y'' = \frac{y(-7) - (-7x)(-\frac{7x}{y})}{y^2}$

Simplifying the expression:

$y'' = \frac{-7y - 49x^2/y}{y^2}$

$y'' = \frac{-7y^2 - 49x^2}{y^3}$

Thus, the second derivative $y''$ is:

$y'' = \frac{-7y^2 - 49x^2}{y^3}$

Summary:

The second derivative $y''$ is:

$y'' = \frac{-7y^2 - 49x^2}{y^3}$

Would you like further details or have any questions?

Here are five related questions:

1. What is implicit differentiation, and how is it applied?
2. How does the quotient rule work for derivatives?
3. What is the geometric meaning of the second derivative?
4. Can we apply this method to other equations with both $x$ and $y$?
5. How does the chain rule affect implicit differentiation?

Tip: The quotient rule is helpful when differentiating ratios of functions; practice using it for implicit differentiation problems!