Explicit functions have 'y' isolated. Implicit functions have 'x' and 'y' mixed, requiring implicit differentiation.

It reveals explicit functions representing parts of the implicit relation and helps determine domain and range.

Positive slope: 'y' increases as 'x' increases. Negative slope: 'y' decreases as 'x' increases. Zero slope: horizontal tangent. Undefined slope: vertical tangent.

The function has a local maximum or minimum, or a point of inflection where the slope is momentarily zero.

The derivative is undefined, often indicating a cusp or a point where the function changes direction sharply.

Differentiate both sides with respect to x, apply the chain rule where necessary, and solve for dy/dx.

The domain of the resulting explicit function(s) may be narrower than the original implicit equation's allowed x-values.

Find dy/dx using implicit differentiation, evaluate at the given point to find the slope, and use the point-slope form of a line.

Related rates problems involve finding the rate at which one quantity is changing by relating it to other quantities whose rates of change are known, often using implicit differentiation.

When differentiating terms involving 'y' with respect to 'x', the chain rule requires multiplying by $\frac{dy}{dx}$ because 'y' is a function of 'x'.

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): $\frac{dy}{dx} = \frac{6-9}{9-6} = -1$.