We are tasked with finding the second derivative $\frac{d^2 y}{dx^2}$ implicitly from the given equation:

$x^2 y - 3x = 6.$

Step 1: Differentiate implicitly to find $\frac{dy}{dx}$

Differentiate both sides of the equation $x^2 y - 3x = 6$ with respect to $x$, remembering to use the product rule on $x^2 y$:

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

Apply the product rule to $x^2 y$ (where $u = x^2$ and $v = y$):

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

The derivative of $-3x$ is $-3$, and the derivative of the constant 6 is 0:

$2x y + x^2 \frac{dy}{dx} - 3 = 0.$

Step 2: Solve for $\frac{dy}{dx}$

Rearrange the equation to isolate $\frac{dy}{dx}$:

$x^2 \frac{dy}{dx} = 3 - 2xy.$

Now, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{3 - 2xy}{x^2}.$

Step 3: Differentiate again to find $\frac{d^2 y}{dx^2}$

Differentiate $\frac{dy}{dx} = \frac{3 - 2xy}{x^2}$ implicitly again with respect to $x$. Use the quotient rule:

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

First, compute $\frac{d}{dx}(3 - 2xy)$. Using the product rule on $-2xy$, we get:

$\frac{d}{dx}(3 - 2xy) = -2y - 2x \frac{dy}{dx}.$

Now plug this back into the quotient rule expression:

$\frac{d^2 y}{dx^2} = \frac{x^2(-2y - 2x \frac{dy}{dx}) - (3 - 2xy)(2x)}{x^4}.$

Simplify the numerator:

$x^2(-2y - 2x \frac{dy}{dx}) - 2x(3 - 2xy) = -2x^2 y - 2x^3 \frac{dy}{dx} - 6x + 4x^2 y.$

The expression becomes:

$\frac{d^2 y}{dx^2} = \frac{-2x^2 y - 2x^3 \frac{dy}{dx} - 6x + 4x^2 y}{x^4}.$

Simplifying further:

$\frac{d^2 y}{dx^2} = \frac{2x^2 y - 6x - 2x^3 \frac{dy}{dx}}{x^4}.$

Finally, substitute $\frac{dy}{dx} = \frac{3 - 2xy}{x^2}$ into the equation:

$\frac{d^2 y}{dx^2} = \frac{2x^2 y - 6x - 2x^3 \left(\frac{3 - 2xy}{x^2}\right)}{x^4}.$

Simplify this:

$\frac{d^2 y}{dx^2} = \frac{2x^2 y - 6x - 2x(3 - 2xy)}{x^4}.$

$\frac{d^2 y}{dx^2} = \frac{2x^2 y - 6x - 6x + 4x^2 y}{x^4}.$

$\frac{d^2 y}{dx^2} = \frac{6x^2 y - 12x}{x^4}.$

Simplify the final expression:

$\frac{d^2 y}{dx^2} = \frac{6y - 12/x}{x^2}.$

This is the second derivative $\frac{d^2 y}{dx^2}$ in terms of $x$ and $y$.

Would you like more details or have any questions?

Related Questions:

1. How does implicit differentiation work for more complex equations?
2. Can you solve a similar problem with different powers of $x$?
3. What is the product rule and how is it used in implicit differentiation?
4. How does the quotient rule help in finding second derivatives?
5. Can you show how to find the third derivative implicitly?

Tip:

Always carefully apply the chain rule when differentiating implicitly, especially when multiple variables are involved!