Calculus Reference

Derivative Definition and Difference Quotient

A derivative measures instantaneous rate of change. It is defined by the limit of the slopes of secant lines as two points move together.

Solved Examples

Examples cover the limit definition, tangent slope, tangent lines, and instantaneous velocity.

Derivative of an expression from the limit definition
Tangent slope from the point definition
Write a tangent-line equation
Find instantaneous velocity

Derivative Definition

At a point

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$

As a function

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

The quotient is the slope of a secant line. Taking the limit as $h\to0$ produces the slope of the tangent line.

Classic Example: $f(x)=x^2$

Use the definition to find the derivative of $f(x)=x^2$.

Difference Quotient

Solution Steps

Substitute $f(x)=x^2$ into the derivative definition.
Expand $(x+h)^2$ and combine like terms.
Factor and cancel the common factor $h$ for $h\ne0$.
Evaluate the remaining limit and state the derivative: $f'(x)=2x$.

$$f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}$$$$=\lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h}$$$$=\lim_{h\to0}\frac{h(2x+h)}{h}$$$$=\lim_{h\to0}(2x+h)=2x$$

At $x=3$, the tangent slope is $f'(3)=2(3)=6$. The derivative gives a new function that reports the slope at every input.

Find a Tangent Slope from the Point Definition

Slope at $x=2$

Use the point definition to find the tangent slope of $f(x)=x^2+1$ at $x=2$.

Solution Steps

Use $a=2$ in the point definition.
Substitute the function values and expand.
Cancel $h$ before taking the limit.
Evaluate the remaining expression at $h=0$.

$$f'(2)=\lim_{h\to0}\frac{[(2+h)^2+1]-[2^2+1]}{h}$$$$=\lim_{h\to0}\frac{4+4h+h^2+1-5}{h}$$$$=\lim_{h\to0}(4+h)$$$$\boxed{f'(2)=4}$$

Write a Tangent-Line Equation

Tangent to a Parabola

Find the equation of the tangent line to $f(x)=x^2+1$ at $x=2$.

Solution Steps

Use the derivative result for the slope at $x=2$.
Find the point on the curve.
Use point-slope form.
Simplify to slope-intercept form.

$$m=f'(2)=4$$$$f(2)=2^2+1=5\quad\Rightarrow\quad(2,5)$$$$y-5=4(x-2)$$$$\boxed{y=4x-3}$$

Find Instantaneous Velocity

Position to Velocity

A particle has position $s(t)=t^2+3t$ meters. Find its instantaneous velocity at $t=2$ seconds.

Solution Steps

Differentiate the position function to obtain velocity.
Substitute the requested time.
State the result with units.

$$v(t)=s'(t)=2t+3$$$$v(2)=2(2)+3=7$$$$\boxed{7\text{ meters per second}}$$

Leibniz Notation

When $y=f(x)$, Leibniz notation writes the derivative as $\frac{dy}{dx}$. Read it as “the derivative of $y$ with respect to $x$.” See the Leibniz notation reference for how to apply the operator, interpret differentials, and use the notation in chain and implicit differentiation.

Geometric and Physical Meaning

Context	Derivative meaning
Graph	Tangent-line slope at a point.
Motion	Instantaneous velocity when position is a function of time.
Data or growth	Instantaneous rate at which one quantity changes as another changes.