Find $\frac{dy}{dx}$ using implicit differentiation, set it equal to 0 and undefined, and solve for x and y.

Positive $\frac{dy}{dx}$ means the function is increasing; negative means decreasing.

Positive $\frac{d^2y}{dx^2}$ means the function is concave up; negative means concave down.

Analyze the sign change of $\frac{dy}{dx}$ around a critical point. Positive to negative indicates a local maximum, negative to positive indicates a local minimum.

If $f''(x) > 0$, the function is concave up. If $f''(x) < 0$, the function is concave down.

It marks a change in the concavity of the function, where the second derivative changes sign.

Find $\frac{dy}{dx}$ and determine the intervals where it is positive (increasing) or negative (decreasing).

Find $\frac{d^2y}{dx^2}$ and determine the intervals where it is positive (concave up) or negative (concave down).

It relates the rates of change of different variables with respect to time, allowing us to find $\frac{dy}{dt}$ given $\frac{dx}{dt}$ and $\frac{dy}{dx}$.

It helps visualize the relationships between variables and identify the equation that relates them.

$\frac{dy}{dx}$

$F(x, y) = 0$

$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$

$a^2 + b^2 = c^2$

$f''(x)$ or $\frac{d^2y}{dx^2}$

$\frac{dy}{dx} = 0$ or $\frac{dy}{dx}$ is undefined

$\frac{d}{dx} [F(x, y)] = 0$

If $f'(x)$ changes from positive to negative at $x=k$, then $f(x)$ has a relative maximum at $x=k$. If $f'(x)$ changes from negative to positive at $x=k$, then $f(x)$ has a relative minimum at $x=k$.

If $f''(x) > 0$, then $f(x)$ is concave up. If $f''(x) < 0$, then $f(x)$ is concave down.

$2x + 2y \frac{dy}{dx} = 0$

It represents the value of $\frac{dy}{dx}$ at that point, indicating the rate of change of y with respect to x.

Look for points where the tangent line is horizontal (local max/min) or vertical (undefined derivative).

The second derivative is positive in that region.

The second derivative is negative in that region.

Look for points where the concavity changes (from concave up to concave down or vice versa).

The derivative $\frac{dy}{dx}$ is undefined at that point.

Implicit functions may not pass the vertical line test, and their graphs can be more complex.

Circle with radius 5 centered at the origin. Top half has positive y values, bottom half has negative y values.

Positive values indicate increasing function, negative values indicate decreasing function, zero values indicate critical points.

Positive values indicate concave up, negative values indicate concave down, zero values indicate potential inflection points.