Question: top-left: Find $$\frac{d^2y}{dx^2}$$, where $$\sqrt{y} + 6xy = 3$$.
bottom-left: $$\frac{d^2y}{dx^2}$$ = User:
1. **State the problem:** We need to find the second derivative $$\frac{d^2y}{dx^2}$$ given the implicit equation $$\sqrt{y} + 6xy = 3$$.
2. **Rewrite the equation:** $$\sqrt{y} + 6xy = 3$$ can be written as $$y^{1/2} + 6xy = 3$$.
3. **Differentiate implicitly with respect to $$x$$:**
$$\frac{d}{dx}\left(y^{1/2}\right) + \frac{d}{dx}(6xy) = \frac{d}{dx}(3)$$
Using the chain rule:
$$\frac{1}{2}y^{-1/2} \frac{dy}{dx} + 6 \left(y + x \frac{dy}{dx}\right) = 0$$
4. **Simplify:**
$$\frac{1}{2} y^{-1/2} \frac{dy}{dx} + 6y + 6x \frac{dy}{dx} = 0$$
Group terms with $$\frac{dy}{dx}$$:
$$\left(\frac{1}{2} y^{-1/2} + 6x\right) \frac{dy}{dx} = -6y$$
5. **Solve for $$\frac{dy}{dx}$$:**
$$\frac{dy}{dx} = \frac{-6y}{\frac{1}{2} y^{-1/2} + 6x}$$
6. **Find the second derivative $$\frac{d^2y}{dx^2}$$:** Differentiate $$\frac{dy}{dx}$$ implicitly using the quotient rule:
Let $$u = -6y$$ and $$v = \frac{1}{2} y^{-1/2} + 6x$$.
Then,
$$\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$
7. **Calculate $$\frac{du}{dx}$$:**
$$\frac{du}{dx} = -6 \frac{dy}{dx}$$
8. **Calculate $$\frac{dv}{dx}$$:**
$$v = \frac{1}{2} y^{-1/2} + 6x$$
Differentiate term by term:
$$\frac{d}{dx} \left(\frac{1}{2} y^{-1/2}\right) = \frac{1}{2} \cdot (-\frac{1}{2}) y^{-3/2} \frac{dy}{dx} = -\frac{1}{4} y^{-3/2} \frac{dy}{dx}$$
$$\frac{d}{dx}(6x) = 6$$
So,
$$\frac{dv}{dx} = -\frac{1}{4} y^{-3/2} \frac{dy}{dx} + 6$$
9. **Substitute back into the formula for $$\frac{d^2y}{dx^2}$$:**
$$\frac{d^2y}{dx^2} = \frac{v (-6 \frac{dy}{dx}) - u \left(-\frac{1}{4} y^{-3/2} \frac{dy}{dx} + 6\right)}{v^2}$$
10. **Substitute $$u$$ and $$v$$ and simplify:**
$$\frac{d^2y}{dx^2} = \frac{\left(\frac{1}{2} y^{-1/2} + 6x\right)(-6 \frac{dy}{dx}) - (-6y) \left(-\frac{1}{4} y^{-3/2} \frac{dy}{dx} + 6\right)}{\left(\frac{1}{2} y^{-1/2} + 6x\right)^2}$$
This expression gives the second derivative $$\frac{d^2y}{dx^2}$$ implicitly in terms of $$x$$, $$y$$, and $$\frac{dy}{dx}$$.
**Final answer:**
$$\boxed{\frac{d^2y}{dx^2} = \frac{\left(\frac{1}{2} y^{-1/2} + 6x\right)(-6 \frac{dy}{dx}) + 6y \left(\frac{1}{4} y^{-3/2} \frac{dy}{dx} - 6\right)}{\left(\frac{1}{2} y^{-1/2} + 6x\right)^2}}$$