$f(x)$ is increasing.

$f(x)$ is decreasing.

$f(x)$ has a horizontal tangent, which could be a local max, local min, or a stationary point.

$f(x)$ is concave up.

$f(x)$ is concave down.

$f'(x) = 0$ (a horizontal line at y=0).

$f'(x)$ does not exist at that point (non-differentiable).

The graph of $f'(x)$ is a horizontal line, representing the constant slope of $f(x)$.

The derivative of a function at a point gives the slope of the tangent line to the function's graph at that point.

The derivative indicates whether the function is increasing or decreasing, and the rate at which it is changing.

The limit allows us to find the instantaneous rate of change by considering what happens as the interval over which we calculate the average rate of change becomes infinitesimally small.

It provides the fundamental basis for understanding and calculating derivatives, and it connects the concept of slope to curves.

It means we are considering the slope of the secant line as the two points get infinitely close together, effectively becoming the tangent line.

A function is differentiable at a point if its derivative exists at that point. This implies the function is continuous and has a well-defined tangent line.

If a function is differentiable at a point, it must be continuous at that point. However, a continuous function is not necessarily differentiable.

The derivative can model various real-world phenomena, such as velocity (rate of change of position), acceleration (rate of change of velocity), and rates of chemical reactions.

A horizontal tangent line indicates that the derivative is zero at that point, implying a local maximum, local minimum, or a stationary point.

$y'$, $f'(x)$, and $\frac{dy}{dx}$ all represent the same concept: the derivative of a function. They are used interchangeably depending on the context.

1. Plug into the limit definition: $\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$. 2. Expand: $\lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$. 3. Simplify: $\lim_{h \to 0} \frac{2xh + h^2}{h}$. 4. Factor and cancel: $\lim_{h \to 0} (2x + h)$. 5. Evaluate the limit: $2x$.

1. Find $f'(x)$: $f'(x) = 2x$. 2. Find the slope at $x = 1$: $f'(1) = 2$. 3. Find the y-coordinate at $x = 1$: $f(1) = 1$. 4. Use point-slope form: $y - 1 = 2(x - 1)$.

1. Find $f'(x)$: $f'(x) = 3x^2 - 3$. 2. Set $f'(x) = 0$: $3x^2 - 3 = 0$. 3. Solve for $x$: $x = \pm 1$.

1. Find a common denominator: $\lim_{h \to 0} \frac{\frac{x - (x+h)}{x(x+h)}}{h}$. 2. Simplify: $\lim_{h \to 0} \frac{\frac{-h}{x(x+h)}}{h}$. 3. Multiply by the reciprocal: $\lim_{h \to 0} \frac{-h}{x(x+h)h}$. 4. Cancel $h$: $\lim_{h \to 0} \frac{-1}{x(x+h)}$. 5. Evaluate the limit: $\frac{-1}{x^2}$.

1. Plug into the limit definition: $\lim_{h \to 0} \frac{[3(x+h)^2 + 4(x+h)] - [3x^2 + 4x]}{h}$. 2. Expand: $\lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 + 4x + 4h - 3x^2 - 4x}{h}$. 3. Simplify: $\lim_{h \to 0} \frac{6xh + 3h^2 + 4h}{h}$. 4. Factor and cancel: $\lim_{h \to 0} (6x + 3h + 4)$. 5. Evaluate the limit: $6x + 4$.