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 $x$.