Question:

Let $f(x)$ be a differentiable function defined implicitly by the equation $x^2 + y^2 - 3xy = 6$. Find $\frac{{dy}}{{dx}}$ at the point $(1,2)$.

Answer:

Given: $x^2 + y^2 - 3xy = 6$

Let's differentiate both sides of the equation with respect to $x$.

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

Using the sum rule and the chain rule, we can differentiate each term on the left side of the equation.

$\frac{{d}}{{dx}}(x^2) + \frac{{d}}{{dx}}(y^2) - \frac{{d}}{{dx}}(3xy) = 0$

Differentiating each term, we get:

$2x + 2yy' - (3y + 3xy') = 0$

$\Rightarrow 2x + 2yy' - 3y - 3xy' = 0$

Now, we can rearrange the equation to solve for $\frac{{dy}}{{dx}}$.

$2yy' - 3xy' = 3y - 2x$

$y'(2y - 3x) = 3y - 2x$

$\Rightarrow y' = \frac{{3y - 2x}}{{2y - 3x}}$

To find $\frac{{dy}}{{dx}}$ at the point $(1,2)$, substitute $x = 1$ and $y = 2$ into the equation we just derived.

$y' = \frac{{3(2) - 2(1)}}{{2(2) - 3(1)}}$

Solving the expression on the right side, we find:

$y' = \frac{{6 - 2}}{{4 - 3}}$

$\Rightarrow y' = \frac{{4}}{{1}}$

Hence, the slope of the curve at the point $(1,2)$ is $\frac{{dy}}{{dx}} = 4$.