Calculus Reference
Derivatives from Limits
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
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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.
$$v(t)=s'(t)=2t+3$$$$v(2)=2(2)+3=7$$$$\boxed{7\text{ meters per second}}$$
Derivative Notation
| Notation | Read as | Meaning |
|---|---|---|
| $f'(x)$ | f prime of x | The derivative of $f$ at $x$. |
| $y'$ | y prime | The derivative when $y=f(x)$. |
| $\frac{dy}{dx}$ | dy over dx | The rate of change of $y$ with respect to $x$. |
| $\frac{d}{dx}[f(x)]$ | d by dx of f of x | An operator that means “differentiate.” |
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. |