As x increases, y also increases. The function is increasing at that point.

As x increases, y decreases. The function is decreasing at that point.

The derivative is zero, indicating a local maximum, local minimum, or a point of inflection.

The derivative is undefined, often indicating a cusp or a sharp turn in the graph.

Domain: the set of all x-values covered by the graph. Range: the set of all y-values covered by the graph.

If replacing x with -x or y with -y leaves the equation unchanged, the graph has symmetry about the y-axis or x-axis, respectively.

Concave up: second derivative is positive. Concave down: second derivative is negative.

Increasing: the graph has a positive slope. Decreasing: the graph has a negative slope.

The function repeats its values in regular intervals. This can be confirmed by checking if $f(x+T) = f(x)$ for some constant T.

The graph is a circle. The derivative gives the slope of the tangent line at any point on the circle. The slope will be undefined at $x = \pm r$, and zero at $y = \pm r$.

An equation where x and y are related, but y is not explicitly isolated.

A function where y is isolated on one side of the equation, expressed directly in terms of x.

The rate of change of y with respect to x (dy/dx) at that point, representing the tangent line's steepness.

An interval where the rate of change of x with respect to y is zero (dx/dy = 0), resulting in a slope of zero.

An interval where the rate of change of y with respect to x is undefined, resulting in an undefined slope.

Isolating y in terms of x, which may result in one or more explicit functions representing parts of the original implicit relation.

The set of all possible x-values for which the function is defined.

The set of all possible y-values that the function can take.

A method used to find the derivative of an implicitly defined function by differentiating both sides of the equation with respect to x, treating y as a function of x.

The graph is the set of all ordered pairs (x, y) that satisfy the equation.

1. Differentiate: $2x + 2y\frac{dy}{dx} = 0$. 2. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{x}{y}$.

1. Find $\frac{dy}{dx} = -\frac{x}{y}$. 2. Evaluate at the point: $\frac{dy}{dx} = -1$. 3. Use point-slope form: $y - \frac{\sqrt{2}}{2} = -1(x - \frac{\sqrt{2}}{2})$.

Try to solve for y. If you get a single expression for y in terms of x, then y is a function of x. If you get multiple expressions, it may not be.

1. Solve for y: $y = \pm \sqrt{4 - x^2}$. 2. Domain: $-2 \le x \le 2$. 3. Range: $-2 \le y \le 2$.

1. Identify variables and rates. 2. Write the equation relating the variables. 3. Differentiate implicitly with respect to time. 4. Substitute known values and solve for the unknown rate.

1. Find $\frac{dy}{dx} = -\frac{x}{y}$. 2. Set $\frac{dy}{dx} = 0$, which implies $x = 0$. 3. Solve for y: $y = \pm 1$.

1. Find $\frac{dy}{dx} = -\frac{x}{y}$. 2. Set the denominator to zero, i.e., $y = 0$. 3. Solve for x: $x = \pm 1$.

1. Find $\frac{dy}{dx}$ using implicit differentiation. 2. Differentiate $\frac{dy}{dx}$ with respect to x, again using implicit differentiation and the quotient rule if necessary. 3. Substitute the expression for $\frac{dy}{dx}$ to simplify.

1. Find $\frac{dy}{dx}$ using implicit differentiation. 2. Evaluate $\frac{dy}{dx}$ at the given point to find the slope of the tangent line. 3. The slope of the normal line is the negative reciprocal of the tangent line's slope. 4. Use the point-slope form of a line to find the equation of the normal line.

1. Differentiate implicitly: $3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$. 2. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2y-x^2}{y^2-2x}$. 3. Evaluate at (3,3): $\frac{dy}{dx} = \frac{6-9}{9-6} = -1$.