Find 
y''
 by implicit differentiation.
6x2 + y2 = 5

To find \( y'' \) by implicit differentiation of the equation \( 6x^2 + y^2 = 5 \), follow these steps:

### Step 1: Differentiate both sides of the equation with respect to \( x \).

We start with:

\[
6x^2 + y^2 = 5
\]

Now, differentiate each term with respect to \( x \), applying the chain rule for \( y^2 \) since \( y \) is a function of \( x \):

\[
\frac{d}{dx}(6x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(5)
\]

### Step 2: Differentiate the terms.

- The derivative of \( 6x^2 \) is:
  \[
  \frac{d}{dx}(6x^2) = 12x
  \]
- The derivative of \( y^2 \) is:
  \[
  \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} = 2y y'
  \]
- The derivative of \( 5 \) is:
  \[
  \frac{d}{dx}(5) = 0
  \]

Thus, we now have the equation:
\[
12x + 2yy' = 0
\]

### Step 3: Solve for \( y' \).

To isolate \( y' \), solve the equation \( 12x + 2yy' = 0 \) for \( y' \):

\[
2yy' = -12x
\]
\[
y' = \frac{-12x}{2y}
\]
\[
y' = \frac{-6x}{y}
\]

### Step 4: Differentiate \( y' \) to find \( y'' \).

Now differentiate \( y' = \frac{-6x}{y} \) with respect to \( x \) to find \( y'' \). Use the quotient rule:

\[
y' = \frac{-6x}{y}
\]
The quotient rule is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = -6x \) and \( v = y \).

Differentiate:
- \( u = -6x \), so \( u' = -6 \)
- \( v = y \), so \( v' = y' \)

Now apply the quotient rule:
\[
y'' = \frac{y(-6) - (-6x)(y')}{y^2}
\]

Substitute \( y' = \frac{-6x}{y} \) into this expression:

\[
y'' = \frac{-6y - (-6x)\left( \frac{-6x}{y} \right)}{y^2}
\]
Simplify the terms:
\[
y'' = \frac{-6y - \frac{36x^2}{y}}{y^2}
\]
\[
y'' = \frac{-6y^2 - 36x^2}{y^3}
\]

Thus, the second derivative is:

\[
y'' = \frac{-6(y^2 + 6x^2)}{y^3}
\]

### Final Answer:
\[
y'' = \frac{-6(y^2 + 6x^2)}{y^3}
\]

---

Would you like further explanation or details on any part of this solution?

Here are 5 related questions:
1. How do you apply the chain rule in implicit differentiation?
2. What is the quotient rule, and how does it apply in differentiation?
3. Can you differentiate other implicit equations, such as \( x^2 + y^2 = r^2 \)?
4. How does implicit differentiation differ from explicit differentiation?
5. What are the applications of implicit differentiation in physics or engineering?

**Tip:** Always use the chain rule when differentiating terms that involve \( y \), as it accounts for \( y \) being a function of \( x \).

Find y'' by implicit differentiation. 6x^2 + y^2 = 5

Discover how to find the second derivative y'' by implicit differentiation of the equation 6x^2 + y^2 = 5. This solution explains key concepts like the chain rule and quotient rule, making it ideal for advanced high school or early college students.

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