Find 
y ″
 by implicit differentiation.
5x3 − 2y3 = 9
y'' =

To find \( y'' \) (the second derivative of \( y \) with respect to \( x \)) using implicit differentiation, let's start with the given equation:

\[
5x^3 - 2y^3 = 9
\]

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

Remember that \( y \) is a function of \( x \), so when differentiating terms with \( y \), we apply the chain rule.

\[
\frac{d}{dx}(5x^3) - \frac{d}{dx}(2y^3) = \frac{d}{dx}(9)
\]

Differentiate each term:

- \( \frac{d}{dx}(5x^3) = 15x^2 \)
- \( \frac{d}{dx}(2y^3) = 6y^2 \cdot \frac{dy}{dx} = 6y^2 y' \)
- \( \frac{d}{dx}(9) = 0 \) (since the derivative of a constant is 0)

So the first derivative equation is:

\[
15x^2 - 6y^2 y' = 0
\]

### Step 2: Solve for \( y' \).

Isolate \( y' \):

\[
15x^2 = 6y^2 y'
\]

\[
y' = \frac{15x^2}{6y^2}
\]

Simplifying:

\[
y' = \frac{5x^2}{2y^2}
\]

### Step 3: Differentiate again to find \( y'' \).

Now, we differentiate \( y' = \frac{5x^2}{2y^2} \) implicitly with respect to \( x \).

Using the quotient rule:

\[
y'' = \frac{(2y^2) \cdot \frac{d}{dx}(5x^2) - (5x^2) \cdot \frac{d}{dx}(2y^2)}{(2y^2)^2}
\]

Differentiate the numerator:

- \( \frac{d}{dx}(5x^2) = 10x \)
- \( \frac{d}{dx}(2y^2) = 4y \cdot y' \)

Substituting these into the equation:

\[
y'' = \frac{(2y^2)(10x) - (5x^2)(4y y')}{(2y^2)^2}
\]

Simplify:

\[
y'' = \frac{20xy^2 - 20x^2y y'}{4y^4}
\]

Substitute \( y' = \frac{5x^2}{2y^2} \) into the equation:

\[
y'' = \frac{20xy^2 - 20x^2y \left(\frac{5x^2}{2y^2}\right)}{4y^4}
\]

Simplify:

\[
y'' = \frac{20xy^2 - 50x^4 / y}{4y^4}
\]

Finally, simplifying further:

\[
y'' = \frac{5x(2y^3 - 5x^3)}{2y^5}
\]

Thus, the second derivative is:

\[
y'' = \frac{5x(2y^3 - 5x^3)}{2y^5}
\]

Would you like a detailed explanation of any specific step or need further clarifications?

Here are 5 relative questions for practice:
1. Can you find the first derivative of \( x^2 + y^2 = 1 \) implicitly?
2. How do you apply implicit differentiation for \( x^2y + xy^2 = 5 \)?
3. What is the second derivative of \( y^2 = 4x \)?
4. Can you solve for \( y' \) in \( x^3 + y^3 = 6xy \)?
5. How would you find \( y' \) if \( 2x^2 + 3y^2 = 12 \)?

**Tip:** When using implicit differentiation, always apply the chain rule to any function of \( y \) involving \( x \), as \( y \) is treated as a function of \( x \).

Find y'' by implicit differentiation. 5x^3 − 2y^3 = 9

Learn how to find the second derivative y'' of the equation 5x^3 − 2y^3 = 9 using implicit differentiation. This step-by-step guide covers key calculus concepts such as the chain rule, quotient rule, and how to differentiate implicitly. Suitable for high school and college-level students.

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