The instantaneous rate of change of a function at a given point.

The derivative of the function $f(x)$, representing the slope of the tangent line at $x$.

The rate of change of a function at a specific instant in time, equivalent to the derivative at that point.

A line that touches a curve at a single point and has the same slope as the curve at that point.

A formal way of defining the derivative as the limit of the difference quotient as h approaches zero.

A notation representing the first derivative of $y$ with respect to the independent variable (usually $x$).

Leibniz's notation for the derivative of $y$ with respect to $x$, representing an infinitesimally small change in $y$ divided by an infinitesimally small change in $x$.

How one quantity changes in relation to another quantity, often represented by the derivative.

The expression $\frac{f(x + h) - f(x)}{h}$, used to calculate the average rate of change of a function over an interval of length $h$.

The slope of the tangent line to the curve at a specific point, representing the instantaneous rate of change at that point.

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: $6x + 4$.

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$y - y_1 = m(x - x_1)$

$m = f'(x_0)$, where $x_0$ is the x-coordinate of the point of tangency.

1. Find $f'(x)$. 2. Evaluate $f'(x_0)$ to find the slope $m$. 3. Use point-slope form: $y - f(x_0) = m(x - x_0)$.