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

Explicit: Differentiate directly. Implicit: Differentiate with respect to x, using chain rule for y terms, then solve for dy/dx. Explicit is easier when possible.

Horizontal: dy/dx = 0. Vertical: dy/dx is undefined. Horizontal indicates a local extremum or inflection point. Vertical indicates a cusp or sharp turn.

Solving for x: expresses x as a function of y. Solving for y: expresses y as a function of x. The choice depends on which is easier and what information you need.

Explicit: Chain rule applied when differentiating a composite function directly. Implicit: Chain rule always applied when differentiating y terms with respect to x.

The domain of the explicit solution can be more restricted than the implicit equation due to square roots or other restrictions.

Explicit: Rates are usually directly given. Implicit: Rates are related through an equation, requiring implicit differentiation with respect to time.

Explicit: Quotient rule used when differentiating a quotient of functions directly. Implicit: Quotient rule may be needed when solving for $\frac{dy}{dx}$ after implicit differentiation.

Explicit: Often simpler to visualize directly. Implicit: Can represent more complex relationships, but may require more analysis to graph.

Implicit: Differentiate $x^2+y^2=r^2$ directly. Explicit: Solve for y first, then differentiate. Implicit is often easier.

Explicit: Find dy/dx directly, then use point-slope form. Implicit: Find dy/dx implicitly, then use point-slope form. Implicit is necessary if you cannot solve for y.

$x^2 + y^2 = r^2$

Differentiate both sides of the equation with respect to x, remembering the chain rule for terms involving y. Then, solve for $\frac{dy}{dx}$.

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\frac{dy}{dx} = -\frac{x}{y}$

$y = \pm \frac{1}{2}\sqrt{16 - x^2}$

$m = \frac{dy}{dx}$

$\frac{d}{dx}f(x,y) = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx}$

$\frac{d}{dx}(y^n) = ny^{n-1} \frac{dy}{dx}$

$y - y_0 = \frac{dy}{dx}|_{(x_0, y_0)} (x - x_0)$

$y = \pm \sqrt{1-x^2}$