To find the second derivative $\frac{d^2y}{dx^2}$ using implicit differentiation, we start from the given equation:

$xy + y^3 = 5.$

First, differentiate both sides of the equation with respect to $x$ to find the first derivative $\frac{dy}{dx}$.

1. Differentiate $xy$ with respect to $x$: $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ (using the product rule).

2. Differentiate $y^3$ with respect to $x$: $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$ (using the chain rule).

So, the first derivative of the equation $xy + y^3 = 5$ is: $x\frac{dy}{dx} + y + 3y^2\frac{dy}{dx} = 0.$

Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx}(x + 3y^2) = -y,$ $\frac{dy}{dx} = \frac{-y}{x + 3y^2}.$

Next, differentiate $\frac{dy}{dx}$ again with respect to $x$ to find the second derivative $\frac{d^2y}{dx^2}$.

Differentiate $\frac{dy}{dx} = \frac{-y}{x + 3y^2}$ implicitly:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-y}{x + 3y^2}\right).$

This involves using the quotient rule:

If $u(x) = -y$ and $v(x) = x + 3y^2$, then

$\frac{d^2y}{dx^2} = \frac{v(x)\frac{du}{dx} - u(x)\frac{dv}{dx}}{v(x)^2}.$

Now, compute $\frac{du}{dx}$ and $\frac{dv}{dx}$:

• $\frac{du}{dx} = -\frac{dy}{dx} = \frac{y}{x + 3y^2}$.
• $\frac{dv}{dx} = 1 + 6y\frac{dy}{dx} = 1 + 6y\left(\frac{-y}{x + 3y^2}\right) = 1 - \frac{6y^2}{x + 3y^2}$.

So, the second derivative becomes:

$\frac{d^2y}{dx^2} = \frac{(x + 3y^2)\left(\frac{y}{x + 3y^2}\right) - (-y)\left(1 - \frac{6y^2}{x + 3y^2}\right)}{(x + 3y^2)^2}.$

Simplifying:

$\frac{d^2y}{dx^2} = \frac{y - y + 6y^3}{(x + 3y^2)^3}.$

Thus, the second derivative is:

$\frac{d^2y}{dx^2} = \frac{6y^3}{(x + 3y^2)^3}.$

Let me know if you need further details or have any questions. Would you like me to go into any specific part of this solution?

Related Questions:

1. How does the product rule apply to implicit differentiation?
2. What is the quotient rule and how is it used in differentiation?
3. Why do we use implicit differentiation for equations involving multiple variables?
4. Can you provide another example of finding the second derivative using implicit differentiation?
5. How does the chain rule apply when differentiating functions implicitly?
6. How can we solve $\frac{dy}{dx}$ directly if the equation was solved for $y$ in terms of $x$?
7. What are some common mistakes to avoid when applying implicit differentiation?
8. How would you apply implicit differentiation to trigonometric functions?

Tip:

When working with implicit differentiation, it’s helpful to remember that every time you differentiate $y$ with respect to $x$, you need to multiply by $\frac{dy}{dx}$.