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Calculus – Study Guide

Derivatives

The derivative of a function is the instantaneous rate of change, or the slope of that function at a particular point. For example, the green line in the picture is tangent to the curve f(x) at the point x = 0.75. The slope of the green line in this picture is the derivative of f(x) at the point x = 0.75.

Notations for the derivative include $f'(x),\ \frac{d}{dx}(f(x)),\ \frac{dy}{dx}.\ and\ y'$ .

We can define the derivative in terms of limits. Suppose we take two points on either side of a point x, and draw the line between them, called a secant line. The slope of the secant line is approximately the slope of the function at point x.

Now move those points closer to x, and draw another secant line. The slope of this line is much closer to the slope of the function at x. If we take the limit of the slope of the secant line as these points move closer and closer together, we get the definition of the derivative at x.

The Limit Definition

$f'(x)=\lim_{\Delta x\to\infty} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

The limit definition is the most basic formula for calculating the derivative of a function. In this formula, f(x) is the function to be differentiated, and $\Delta x$ represents a small change in x, which goes to zero.

Example:

Find the derivative of $f(x)=x^{2}$ . Then calculate the slope of $x^2$ at $x=4$ .

$f'(x)=\lim_{\Delta x\to\infty} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

$f'(x)=\lim_{\Delta x\to\infty} \frac{(x+\Delta x)^{2}-x^2}{\Delta x}$ ⇒ Change $f(x)$ to $x^2$ .

$f'(x)=\lim_{\Delta x\to\infty} \frac{x^{2}+2x\cdot\Delta x+(\Delta x)^{2}-x^{2}}{\Delta x}$ ⇒ FOIL the $(x+\Delta x)^{2}$ .

$f'(x)=\lim_{\Delta x\to\infty} \frac{2x\cdot\Delta x+(\Delta x)^{2}}{\Delta x}$ ⇒ The $x^{2}$ terms cancel.

$f'(x)=\lim_{\Delta x\to\infty} 2x+\Delta x$ ⇒ The $\Delta x's$ cancel.

$f'(x)=2x$ ⇒ $\Delta x$ goes to zero.

To find the slope of f(x) at the point x = 4, we plug in 4 to the derivative function.

$f'(x)=2x$
$f'(4)=2(4)=8$

The slope of the line that is tangent to f(x) when x = 4 is 8.

Sometimes, it is not possible to find the derivative of a function at a certain point because that function is not continuous.

Example: Find the derivative of |x| at x = 0.

We can do this one using the limit definition (try it for practice!), or we can look at it graphically:

The absolute value function has a derivative of ±1, depending on the sign of x.

$f(x) = \left\{ \begin{array}{ll} -x & \quad x \leq 0 \\ x & \quad x > 0 \end{array} \right.$

But what is the derivative at x = 0?
When you studied limits, you probably learned that when the right-hand and left-hand limits at a point don’t agree, the limit at that point does not exist.

The derivative of |x| does not exist at x = 0 because the function is not continuous there.

Rules for Finding Derivatives

In many case, the limit definition of the derivative that you will use most often, so you should make yourself familiar with it.

For any function $a\cdot x^n$ , the derivative can be found as:

$\frac{d}{dx}a\cdot x^{n}=a\cdot n\cdot x^{n-1}$

To apply this rule, multiply the coefficient, a, by the current exponent, n, and then decrease the exponent on x by 1.

Examples:

$\frac{d}{dx}(x^{2})=2x$ ⇒ Multiply by 2, then decrease the exponent to 1.
$\frac{d}{dx}(2x^{5})=10x^4$ ⇒ Multiply by 5, then decrease the exponent to 4.
$\frac{d}{dx}(2x^{-2})=-4x^{-3}$ ⇒ Multiply by -2, then decrease the exponent to -3.

The Constant Rule

The derivative of any constant is zero.

This can be observed graphically, because a line representing a constant function is horizontal, with zero slope. This can also be shown in terms of the power rule, because a constant a can be written as $ax^{0}$ .

Example: $\frac{d}{dx}(5)=\frac{d}{dx}(5x^{0})=0$

Derivative of a Sum

The derivative of a sum of terms is equal to the sum of the derivatives of the terms.

$\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)$ .

This makes it easy to differentiate polynomials using the power rule.

Example: $\frac{d}{dx}(2x^{3}+x^{7})=\frac{d}{dx}(2x^{3})+\frac{d}{dx}(x^{7})=6x+7x^6$

Derivatives of Trigonometric Functions

The derivatives of common trigonometric functions should be memorized.

$\frac{d}{dx}(sin(x))=cos(x)$

$\frac{d}{dx}(cos(x))=-sin(x)$

$\frac{d}{dx}(tan(x))=sec^{2}(x)$

$\frac{d}{dx}(sec(x))=sec(x)\cdot tan(x)$

$\frac{d}{dx}(cot(x))=-csc^{2}(x)$

$\frac{d}{dx}(csc(x))=-csc(x)\cdot cot(x)$

The Constant Multiple Rule

If a function is multiplied by a constant coefficient, you may take the derivative of the function alone, and then multiply by the constant.

$\frac{d}{dx}(a\cdot f(x))=a\cdot f'(x)$

Example: $\frac{d}{dx}(3\cdot sin(x))=3\cdot\frac{d}{dx}(sin(x))=3\cdot cos(x)$

Differentiation Rules of Logarithmic and Exponential Functions

These commonly used derivatives should be memorized. Let a be a constant.

$\frac{d}{dx}(e^{x})=e^x$

$\frac{d}{dx}(a^{x})=a^{x}\cdot ln(a)$

$\frac{d}{dx}(ln(x))=\frac{1}{x}$

$\frac{d}{dx}(log_{a}(x))=\frac{1}{x\cdot ln(a)}$

The Product Rule

For two functions multiplied by one another, the product rule can be used:

$\frac{d}{dx}(f(x)\cdot g(x))=f(x)\cdot g'(x)+f'(x)\cdot g(x)$
To apply this rule, differentiate each function separately. Multiply the first function by the derivative of the second, and vice-versa. Then add the two two terms together.

Example: Find the derivative of $f(x)=x^{2}sin(x)$ .

$Let\ u=x^{2}.\ v=sin(x).\ Then\ u'=2x,\ v'=cos(x)$

$f'(x)=u\cdot v'+u'\cdot v=x^{2}cos(x)+2xsin(x)$ .

The Quotient Rule

For two terms divided by one another, the quotient rule is a convenient shortcut.

$\frac{d}{dx}(\frac{f(x)}{g(x)}}=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$
Again, differentiate each function separately. Then multiply the numerator function by the derivative of the denominator and vice versa. Subtract the two terms, as shown. Divide by the square of the denominator.

Example: Find the derivative of $F(x)=\frac{cos(x)}{2x}$

$Let\ f=cos(x).\ g=2x.\ Then\ f'=-sin(x),\ g'=2$

$F'(x)=\frac{g\cdot f'+g'\cdot f}{g^2}=\frac{2xcos(x)-2sin(x)}{4x^2}=\frac{xcos(x)-sin(x)}{2x^2}$

The Chain Rule

For a composition of two function, f(x) and g(x), use the chain rule.

$\frac{d}{dx} f(g(x))=f'(g(x))\cdot g'(x)$
To use this formula, first find the derivative of the inside and outside functions separately. Apply the derivative of the outside function to the inside function. Then multiply by the derivative of the inside function.

Example: Find the derivative of $(2x+1)^3$

Inside function: $2x+1 \Rightarrow \frac{d}{dx}(2x+1)=2$

Outside function: $y^{3} \Rightarrow \frac{d}{dx}(y^{3})=3y^2$

$\frac{d}{dx}((2x+1)^{3})=3(2x+1)^{2}\cdot 2=6(2x+1)^2$

Differentiation Rules of Inverse Trigonometric Functions

It can be useful to memorize these less common derivatives, especially in preparation for finding integrals in the next semester of calculus. You should learn to recognize them.

$\frac{d}{dx}(sin^{-1}(x))=\frac{1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}(cos^{-1}(x))=\frac{-1}{\sqrt{1-x^{2}}}$

$\frac{d}{dx}(tan^{-1}(x))=\frac{1}{1+x^{2}}$

$\frac{d}{dx}(cot^{-1}(x))=\frac{-1}{1+x^{2}}$

$\frac{d}{dx}(sec^{-1}(x))=\frac{1}{|x|\sqrt{x^{2}-1}}$

$\frac{d}{dx}(csc^{-1}(x))=\frac{-1}{|x|\sqrt{x^{2}-1}}$

Basics of Differentiation Practice Problems:

1. Find, using the limit definition, the derivative of x + 2.
2. Find the derivative of:
  - a. $x^{2}$
  - b. $2x^{-5}$
  - c. $3x^{\frac{2}{3}}$
  - d. $2x^{4}-x^{3}+5x$
  - e. $5\ (Think\ of\ this\ as\ 5x^{0}.)$
3. Use the product rule to find the derivative of:
  - a. $(x+3)(x+4)$
  - b. $x^{3}(x^{2}+3x)$
4. Use the quotient rule to find the derivative of:
  - a. $\frac{7x+3}{x+2}$
  - b. $\frac{1}{x^{-2}}$
5. Use the chain rule to find the derivative of:
  - a. $(7x+1)^3$
  - b. $3(x^{3}+6)^{2}+2$

Practice Problem Solutions:

Finding the Equation of the Tangent Line

Now that we know how to find the slope of a function, we can find the equation of the tangent line. Recall that to write the equation of a line we need the slope and a point on the line. We can get the slope from taking the derivative of the function, and we can get a point by plugging it into the function.

Example: Find the equation of the tangent line to the function $f(x)=x^{3}+3x-6$ at the point $x=2$ .

First, find the derivative of the function.

$f'(x)=\frac{d}{dx} (x^{3}+3x-6)$
$f'(x)=\frac{d}{dx} (x^{3})+\frac{d}{dx} (3x)-\frac{d}{dx} (6)$
$f'(x)=3x^{2}+3$

To find the slope at 2, plug $x=2$ into the derivative.

$f'(2)=3\cdot (2^{2})+3$
$f'(2)=15$

Plug $x=2$ into the function to get the y-value of the point.

$f(2)=2^{3}+3(2)-6=8$

Now we have slope $m=15$ at the point (2, 8).
We can use this information to write the equation of the line in point-slope form.

$y-y_{1}=m\cdot (x-x_{1})$
$y-8=15\cdot (x-2)$
$y=15x-22$

We can now plot $f(x)$ and the tangent line $y=15x-22$ on the same set of axes.

Implicit Differentiation

Sometimes it is necessary to find the derivative of a function which is not in the form $y=f(x)$ .
For example:

$x^{2}y-3x=sin(x)$

We cannot just take the derivative of this function because it is not solved for y. Therefore, we will use implicit differentiation. Here are the rules for implicit differentiation:

1. Take the derivative of both sides. Use all of the same differentiation rules as when taking the derivative of an expression involving only x.
2. When you encounter y, take the derivative using all the same rules, but also multiply by $\frac{dy}{dx}$ . (The reason for this lies in the chain rule. y is a function of x.)
3. Once you have taken all of the derivatives, solve for $\frac{dy}{dx}$ .

Let’s find the derivative of the function above.

$x^{2}y-3x=sin(x)$
$x^{2}\cdot 1\frac{dy}{dx}+y\cdot 2x-3=cos(x)$ ⇒ On the left, we have used the product rule.
$\frac{dy}{dx}=\frac{cos(x)-2xy+3}{x^2}$ ⇒ Solve for $\frac{dy}{dx}$

Practice: Try using implicit differentiation on the following functions:

1. $x^{2}+y^{2}=25$
2. $sin(x+y)=y^{2}cos(x)$

Solutions:

1. $x^{2}+y^{2}=25$
  $2x+2y\frac{dy}{dx}=0$ ⇒ Power Rule and Implicit Differentiation.
  $\frac{dy}{dx}=\frac{-x}{y}$ ⇒ Solve for $\frac{dy}{dx}$
2. $sin(x+y)=y^{2}$
  $cos(x+y)\cdot (1+\frac{dy}{dx})=2y\cdot\frac{dy}{dx}$ ⇒ Implicit Differentiation
  $cos(x+y)+\frac{dy}{dx}\cdot cos(x+y)=2y\cdot\frac{dy}{dx}\cdot cos(x)$ ⇒ Distribute $cos(x+y)$ .
  $\frac{dy}{dx}\cdot cos(x+y)-2y\cdot\frac{dy}{dx}\cdot cos(x)=-cos(x+y)$ ⇒ Rearrange the terms.
  $\frac{dy}{dx} (cos(x+y)-2y\cdot cos(x))=y^{2}\cdot (-sin(x))-cos(x+y)$ ⇒ Factor out $\frac{dy}{dx}$ .
  $\frac{dy}{dx}=\frac{-y^{2}sin(x)-cos(x+y)}{cos(x+y)-2y\cdot cos(x)}$ ⇒ Solve for $\frac{dy}{dx}$

Higher Order Derivatives

So far, we have only taken the first derivative. There are also second, third, fourth, and even higher order derivatives. We already know that the first derivative gives us the slope of the function. The second derivative describes the rate of change of the slope. We call this concavity.
Notations for the second derivative includes $f''(x),\ y'',\ or\ \frac{d^{2}y}{dx^{2}}$ .

To find higher order derivatives, you take derivatives successively. For example, to find the second derivative you would first take the first derivative of the function, and then take the derivative again. Let’s do an example.

$f(x)=x^{3}+cos(x)$

$f'(x)=f'(x^{3}+cos(x))$ ⇒ Take first derivative

$f'(x)=3x^{2}-sin(x)$ ⇒ Power Rule

$f''(x)=f'(3x^{2}-sin(x))$ ⇒ Take second derivative

$f''(x)=6x-cos(x)$ ⇒ Power and Trig Rule

Practice: Find the second and third derivatives of the following functions:

1. $cos(2x)$
2. $ln(x)+x^3$
3. $2e^x$

Solutions:

1. $f'(x)=f'(cos(2x))$ ⇒ take first derivative
  $f'(x)=-sin(2x)\cdot 2$ ⇒ Chain Rule
  $f'(x)=-2sin2(2x)$ ⇒ Rewrite
  $f''(x)=f'(-2sin(2x))$ ⇒ Take second derivative
  $f''(x)=-2cos(2x)\cdot 2$ ⇒ Chain Rule
  $f''(x)=-4cos(2x)$ ⇒ Simplify
  $f'''(x)=f'(-4cos(2x))$ ⇒ Take third derivative
  $f'''(x)=-4\cdot (-sin(2x))\cdot 2$ ⇒ Chain Rule
  $f'''(x)=8sin(2x)$
2. $f'(x)=f'(ln(x)+x^{3})$ ⇒ Take first derivative
  $f'(x)=\frac{1}{x}+3x^2$ ⇒ Logarithmic and Power Rule
  $f'''(x)=f'(\frac{1}{x}+3x^{2})$ ⇒ Take second derivative
  $f''(x)=\frac{-1}{x^2}+6x$ ⇒ Power Rule
  $f'''(x)=f'(\frac{-1}{x^2}+6x)$ ⇒ Take third derivative
  $f'''(x)=\frac{2}{x^3}+6$ ⇒ Power Rule
3. $f'(x)=f'(2e^{x})$ ⇒ Take first derivative
  $f'(x)=2e^x$ ⇒ Exponential Rule
  $f''(x)=f'(2e^{x})$ ⇒ Take second derivative
  $f''(x)=2e^{x}$ ⇒ Exponential Rule
  $f''(x)=f'(2e^{x})$ ⇒ Take third derivative
  $f'''(x)=2e^x$ ⇒ Exponential Rule

Critical Points and the Derivative Test

Knowing the first and second derivatives can reveal valuable information about a function. Most importantly, we can determine where the function is increasing, decreasing, concave up, and concave down. Critical points are points where the function changes from increase to decreasing (this is called a maximum), decreasing to increasing (this is a minimum), or changes in concavity (this is called a point of inflection).

First Derivative Test

To find maximums and minimums, take the first derivative and set it equal to zero. Solve for x. These are your first critical points.

Example:

$f(x)=3x^{2}-4x-3$
$f'(x)=9x^{2}-4$
$9x^{2}-4=0$
$x=\pm\sqrt{\frac{4}{9}}$
$x=\pm\frac{2}{3}$

The first two critical points are $\frac{-2}{3}$ and $\frac{2}{3}$ . But we don’t yet know whether these are minimum or maximums. We need to examine the first derivative to figure that out. To do this, imagine or sketch a number line with the critical points labeled as shown in the figure below.

Next, choose a point for each of the three ranges. We want to see if the derivative is positive or negative in each range. This will tell us if the function is increasing (positive slope) or decreasing (negative slope).

From the above picture, you can see that -1, 0, and 1 will be good choices. Now, evaluate $f'(x)$ for each of the values x = -1, x = 0, and x = 1.

$f'(-1)=9(-1)^{2}-4$
$f'(-1)=9-4$
$f'(-1)=5$
Positive (+)

$f'(0)=9(0)^{2}-4$
$f'(0)=0-4$
$f"(0)=-4$
Negative (-)

$f'(1)=9(1)^{2}-4$
$f'(1)=9-4$
$f'(1)=5$
Positive (+)

Evaluating $f'(x)$ we see that the slope of the graph is positive, then negative, and then positive. This tells us that the first critical point is a maximum, and the second critical point is a minimum. Here is a picture of this function:

Second Derivative Test

We can do the same things with the second derivative that we did with the first derivative. This time the critical points will be points of inflection, and a positive second derivative will indicate concave up and a negative second derivative will indicate concave down. Let’s continue the above problem by taking the second derivative and setting it equal to zero.

$f''(x)=f'(9x^{2}-4)$

$f''(x)=18x$

$18x=0$

$x=0$

Our critical point is x = 0. Let’s set up a number line so we can choose a point in each important range.

We will use -1 for range 1 and 1 for range 2. Evaluate the second derivative at each of these points.

$f''(x)=18x$
$f''(-1)=18\cdot -1$
$f''(-1)=-18$
Negative (-)

$f''(x)=18x$
$f''(1)=18\cdot 1$
$f"'(1)=18$
Positive (+)

This tells us that the function is concave down from $-\infty$ to 0, and concave up from 0 to $\infty$ . Looking at the graph of the function, we can see this is correct.

Example:
find the intervals of increasing, decreasing, concave up, and concave down for

$f(x)=cos(x)$

On the interval $(-2\pi,\ 2\pi)$ .

First let’s find the critical points for the first derivative:

$f'(x)=f'(cos(x))$
$f'(x)=-sin(x)$
$-sin(x)=0$
$x=0, \pi ,\ -\pi$

Now, set up a number line.

We will evaluate the first derivative at $x=\frac{-3\pi}{2},\ \frac{-\pi}{2},\ \frac{\pi}{2},\ and\ \frac{3\pi}{2}$ .

$f'(x)=-sin(x)$

$f'(\frac{-3\pi}{2})=-sin(\frac{-3\pi}{2})$
$f'(\frac{-3\pi}{2})=-1$
Negative (-)

$f'(\frac{-\pi}{2})=-sin(\frac{-\pi}{2})$
$f'(\frac{-\pi}{2})=1$
Positive (+)

$f'(\frac{\pi}{2})=-sin(\frac{\pi}{2})$
$f'(\frac{\pi}{2})=-1$
Negative (-)

$f'(\frac{3\pi}{2})=-sin(\frac{3\pi}{2})$
$f'(\frac{3\pi}{2})=1$
Positive (+)

-π is a minimum, 0 is a maximum, and π is a minimum
This function is increasing on the interval (-π, 0) ∪ (π. 2π) and decreasing on the interval (-2π, -π) ∪ (0, π).

Check by graphing:

Now, lets do the second derivative test. To do this, evaluate the second derivative and set it equal to zero to find the critical points.

$f''(x)=f'(-sin(x))$
$f''(x)=-cos(x)$
$-cos(x)=0$

The critical points are $x=-\frac{3\pi}{2},\ -\frac{\pi}{2},\ \frac{\pi}{2},\ \frac{3\pi}{2}$ .

We will evaluate the second derivative at -π, 0, and π.

$f''(x)=-cos(x)$
$f''(-\pi)=-cos(-\pi)$
$f''(-\pi)=1$
Positive (+)

$f''(0)=-cos(0)$
$f''(0)=-1$
Negative (-)

$f''(\pi)=-cos(\pi)$
$f''(\pi)=1$
Positive (+)

This function is concave up on the interval $(-\frac{3\pi}{2},\ -\frac{\pi}{2})$ , concave down on the interval $(-\frac{\pi}{2},\ \frac{\pi}{2})$ and concave up on the interval $(\frac{\pi}{2},\ \frac{3\pi}{2})$ .

Check by graphing:

Limits

The limit of a function $f(x)$ at a value c is the value that $f(x)$ approaches as x gets closer to c.

Notation

$\lim_{x\to\ c} f(x)$ ⇒ “The limit as x approaches c of f(x)”

Formal Definition

$\lim_{x\to\ c} f(x)=L$ if an only if for every ε > 0, there exists δ such that 0 < |x – c| < δ implies |f(x) – L| < ε.

Direct Substitution

Some limits can be evaluated by simply substituting c for x in the function f(x).

Example: $\lim_{x\to\ 3} x^{2}=3^{2}=9$

Direct substitution works anytime f(x) is continuous and f(c) can be calculated. This applies to constant function, polynomial functions, and rational functions whenever c is not infinity or zero.

However, limits are more often used when f(x) is discontinuous or f(c) is undefined.

Approximating Limits

A limit can often be estimated by making a chart of value of f(x) for values of x which approach c. If the values of f(x) approach a number, it is a good guess for the limit.

Example: $\lim_{x\to\ c} \frac{x^{3}-1}{x-1}$

We can also obtain this result graphically.

Although the function is discontinuous at = 1, we can see that f(x) approaches 3 when x is close to 1.

One-sided Limits

For discontinuous functions (such as some piecewise functions), the limits at a point may not exist. Instead, you can use a right-hand or left-hand limit.

Example:

$for\ x<1\rightarrow f(x)=x$
$for\ x\geq 1\rightarrow f(x)=x^{2}-1$

Here, the limit at 1 takes on two different values, depending on whether you approach x = 1 from the left (smaller values) or from the right (larger values).

Right limit: $\lim_{x\to\ 1+} f(x)=0$ Left limit: $\lim_{x\to\ 1-} f(x)=1$

When $\lim_{x\to\ c+} f(x)\neq\lim_{x\to\ c-} f(x)$ , we say that $\lim_{x\to\ c} f(x)$ does not exist.

If $\lim_{x\to\ c+} f(x)=\lim_{x\to\ c-} f(x)$ , then $\lim_{x\to\ c} f(x)$ exists.

Limits that approach $\frac{a}{0},\ \frac{0}{0},\ or\ \pm\frac{\infty}{\intfy}$ .

The limit of a rational function where the numerator approaches some positive value and the denominator approaches zero is infinity. If the numerator approaches a negative value, and the denominator approaches zero, the limit is negative infinity.

Examples: $\lim_{x\to\ 0} \frac{x+2}{3x}=\infty$ $\lim_{x\to\ 0} \frac{x-2}{3x}=\infty$

Often you will be asekd to take the limit of a rational function f(x) = p(x)/q(x) where p(x) and q(x) both approach zero of infinity. Here are some tips for finding these limits:

- Try to simplify the expression.Example: $\lim_{x\to\ 2} \frac{3x-6}{2x-4}=\lim_{x\to\ 2} \frac{3(x-2)}{2(x-2)}=\lim_{x\to\ 2} \frac{3}{2}=\frac{3}{2}$
- Take the derivative of the top and bottom, and use L’Hopital’s rule:If p(x), q(x) = 0 or $\pm\infty$ . then $\lim_{x\to\ c} \frac{p(x)}{q(x)}=\lim_{x\to\ c} \frac{p'(x)}{q'(x)}$ Example: $\lim_{x\to\ 0} \frac{2x}{3x^2}=\lim_{x\to\ 0} \frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(3x^{2})}=\lim_{x\to\ 0} \frac{2}{6x}=\infty$

Limits at Infinity

Often you will be asked to take the limit of a function as x approaches infinity.

When taking the limit at infinity of a rational function f(x) = p(x)/q(x) where p(x) and q(x) are polynomials:

$\lim_{x\to\ \infty} \frac{p(x)}{q(x)}$

- If the degree of p is greater than the degree of q, then the limit is positive or negative infinity depending on the signs of the leading coefficients;
- If the degree of p and q are equal, the limit is the leading coefficient of p divided by the leading coefficient of q;
- If the degree if p is less than the degree of q, the limit is 0.

Examples:

- $\lim_{x\to\ \infty} \frac{7x^{3}+2x^{2}+1}{3x^3}=\frac{7}{3}$
- $\lim_{x\to\ 0} \frac{-5x^2}{3x+1}=-\infty$
- $\lim_{x\to\ 0} \frac{x^{2}+3x}{3x^5}=0$

Properties of Limits

Let b and c be real numbers, and let n be a positive integer.

1. Scalar multiple: $\lim_{x\to\ c} b\cdot f(x)=b\cdot\lim_{x\to\ c} f(x)$
2. Sum or difference: $\lim_{x\to\ c} [f(x)\pm g(x)]=\lim_{x\to\ c} f(x)\pm\lim_{x\to\ c} g(x)$
3. Product: $\lim_{x\to\ c} [f(x)\cdot g(x)]=\lim_{x\to\ c} f(x)\cdot\lim_{x\to\ c} g(x)$
4. Power: $\lim_{x\to\ c} [f(x)]^{n}=[\lim_{x\to\ c} f(x)]^{n}$

Practice Problems

Find each limit using direct substitution.
- $\lim_{x\to\ 3} x^{2}+3x+1$
- $\lim_{x\to\ 0} 27$
- $\lim_{x\to\ 0} \frac{1}{x}$
Find a right-hand and left-hand limit as x approaches 2 for the function
- if x ≤ 2 → $f(x)=3x^{2}-5$
- if x > 2 → $f(x)=2x+4$
  Then determine whether the limit $\lim_{x\to\ 2} f(x)$ exists.
Simplify each expression to find the limit.
- $\lim_{x\to\ 3} \frac{-27x+9}{9x-3}$
- $\lim_{x\to\ 2} \frac{3x-6}{x^{2}-4}$
Find each limit using L’Hopital’s rule.
- $\lim_{x\to\ \infty} \frac{x^2}{3x+5}$
- $\lim_{x\to\ 0} \frac{e^{x}-1}{5x}$
Find each limit at infinity using the lead terms of the polynomials.
- $\lim_{x\to\ \infty} \frac{x^2}{3x+5}$
- $\lim_{x\to\ \infty} \frac{x^{7}+2x^{5}+x^{2}+9}{3x^{7}+5x^{4}+24}$
- $\lim_{x\to\ \infty} \frac{8x^{2}+6}{3x^{4}+1}$

Solutions to Practice Problems

Riemann Sums

Introduction to Riemann Sums:
Riemann sums is a method for approximating the total area underneath a curve on a graph, also known as an integral.

This is a graph of the line $y=x^2$

Suppose we want to find the area under the curve for the interval x = 2 to x = 8 (starting at 2, ending at 8 along the x-axis)

We can split the graph into a finite number of rectangles, and use these to approximate the area.

As the rectangles become smaller we get a better estimation of the area under the curve. (The true area in this example is 168.)

How to Calculate Riemann Sums:

Step 1: Find the width of the rectangles: $width\ of\ rectangles=\frac{b-a}{n}$

a = the number you start at
b = the number you end with
n = number of rectangles you want to approximate with.

Step 2: Find the heights of the rectangles: $height\ of\ rectangles=f(x)$

Each rectangle spans a range of x-values. Which do we plug in to calculate the height?

Left-hand Estimation: Use the x-value at the left end of each rectangle to calculate height.
Right-hand Estimation: Use the x-value at the right of each rectangle to calculate height.
Midpoint Estimation: Use the x-value in the middle of each rectangle to calculate height.

Calculate f(x) at each left, right or midpoint value to find the heights.

Note: For increasing functions, a left-hand sum will give an under-estimate, and a right-hand sum will give an over-estimate. For decreasing functions, a left-hand sum will give an over-estimate, and a right-hand sum will give an under-estimate.

Step 3: Find the areas of all the rectangles by multiplying width times height. Add these together to estimate the area under the curve.

Example Problem:

Use four rectangles to find an upper and lower approximation of the area under $f(x)=20-x^{2}$ between x = -2 and x = 4.

First find the width of each rectangle. To do this, divide the total width of the interval by the number of rectangles you want.

$\frac{b-a}{n}=\frac{4-(-2)}{4}=\frac{6}{4}=\frac{3}{2}$

Next we need the heights of the rectangles. To find the right-hand sum we need to use the heights at the right endpoints of each rectangle.

The right endpoints of the four rectangles are $-\frac{1}{2},\ 1,\ 2\frac{1}{2},\ 4$
The function is $f(x)=20-x^2$ .

Height of Rectangle 1 = $f(-0.5)=20-(-0.5)^{2}=19.75$
Height of Rectangle 2 = $f(1)=20-(1)^{2}=19$
Height of Rectangle 3 = $f(2.5)=20-(2.5)^{2}=13.75$
Height of Rectangle 4 = $f(4)=20-(4)^{2}=4$

Add up the areas of all the rectangles. $A=l\times w$

$A=(1.5\times 19.75)+(1.5\times 19)+(1.5\times 13.75)+(1.5\times 4)$
$=29.625+28.5+20.625+6=84.75$

Note: Since all of the rectangles are of equal width, you could sum the heights first, and then multiply once by the width.

To find the left-hand sum you will need to use the heights at the left endpoints of each rectangle.

The left endpoints of the four rectangles are $-2,\ -\frac{1}{2},\ 1,\ 2\frac{1}{2}$
The function is $f(x)=20-x^2$

Height of Rectangle 1 = $f(-2)=20-(-2)^{2}=16$
Height of Rectangle 2 = $f(-0.5)=20-(-0.5)^{2}=19.75$
Height of Rectangle 3 = $f(1)=20-(1)^{2}=19$
Height of Rectangle 4 = $f(2.5)=20-(2.5)^{2}=13.75$

Sum the area of all the rectangles $A=l\times w$

$A=(1.5\times 16)+(1.5\times 19.75)+(1.5\times 19)+(1.5\times 13.75)$
$=24+29.625+28.5+20.625=102.75$

In this example, the right-hand sum is a lower approximation. The left-hand sum is an upper approximation. We can conclude that the true area is between 84.75 and 102.75.

Finding the Exact Area through Integrals:

In this integral, a and b are the upper and lower boundaries. The function f(x) is the integrand. The dx at the end tells us that we are taking the integral with respect to the variable x.

When estimating the area under a curve using Riemann sums, we saw that using more rectangles led to more accurate approximations. If we let the size of the rectangles get smaller and smaller, and the number of rectangles get larger and larger, the sum of the rectangles approach the true area under the curve. Taking the limit of a Riemann approximation as n (the number of rectangles) goes to infinity gives the exact area under the curve, also called a definite integral.

$Exact\ Area\ =\int_{a}^{b} f(x)\ dx=\lim_{n\to\ \infty}[(\frac{b-a}{n})\cdot\sum_{k=1}^{n} f(a+k(\frac{b-a}{n}))]$

This integral is shown as the limit of a right-hand sum. A left-hand sum or midpoint sum would have the same limit as n approaches infinity, and would give the same integral.

Definite and Indefinite Integrals

Integration is the opposite of differentiation. For this reason, the terms “integral” and “anti-derivative” are sometimes interchanged. Integrals, like derivatives, can be calculated without taking a limit.

An integral of the form $\int_{a}^{b} f(x)\ dx$ , with upper and lower bounds, is called a definite integral. It represents the area under the curve $f(x)$ from a to b, and it is equal to a number.

An integral of the form $\int f(x)\ dx$ , without upper and lower bounds, is called an indefinite integral. It represents all functions that have the derivative $f(x)$ , and is euqal to a family of curves.

$\int f'(x)\ dx=f(x)+c$

Since the derivative of a constant is zero, taking the anti-derivative of $f'(x)$ produces a family of curves, $f(x)+c$ , where $c$ can by any constant.

Example: What is $\int 2x\ dx$ ?

We know the derivative of $x^{2}$ is $2x$ . But the derivatives of $x^{2}+1,\ x^{2}+7,\ and\ x^{2}+69$ are also all equal to $2x$ . In fact, the derivatives of $x^2$ plus any number is $2x$ .

$\int 2x\ dx=x^{2}+c \Rightarrow where\ c\ is\ any\ constant$

It is imporant to remember to add c when taking an indefinite integral.

The Fundamental Theorem of Calculus

Definite integrals can be calculated from indefinite integrals using The Fundamental Theorem of Calculus.

$\int_{a}^{b} f(x)\ dx=F(b)-F(a)$

Where $\inf f(x)=F(x)+c$ . (Notice that the $+c$ doesn’t matter here, because when we subtract $(F(b)+c)-(F(a)+c)$ , the $c$ ‘s will cancel out.)

Example: Find the area under the curve $f(x)=3x^{2}$ from x = 1 to x = 2.

$\int_{1}^{2} 3x^{2}\ dx\ \Rightarrow\ Set\ up\ the\ integral.$

$\int 3x^{2}\ dx=x^{3}\ \Rightarrow\ Use\ the\ reverse\ power\ rule\ to\ find\ an\ antiderivative.$

$\int_{1}^{2} 3x^{2}\ dx=(2)^{3}-(1)^{3}=7\ \Rightarrow\ Evaluate\ the\ definite\ integral.$

Methods for Integration

When taking the integral of a function, you must apply the rules of differentiation in reverse. This can be tricky at first.

Example: Apply the reverse power rule to find the following indefinite integral:

$\int 2x^{3} dx$ First, think of a derivative that has an $x^3$ in it. I know that $\frac{d}{dx} (x^{4})=4x^3$ .
We need to turn that 4 into a 2, so introduce a factor of $\frac{1}{2}$ .
Deduce that $\frac{d}{dx} (\frac{1}{2} x^{4})=2x^3$ .
Write the anti-derivative, and add c.

$\int 2x^{3} dx=\frac{1}{2} x^{4}+c$

U-substitution:

You should recall the chain rule from differentiation. When the chain rule is applied to a function $f(g(x))$ , the derivative has the form $f'(g(x))\cdot g'(x)$ .

When integrating a function of the form $f'(g(x))\cdot g'(x)$ , we do the chain rule in reverse.

First, make the substitution $u=g(x)$ . Find $\frac{du}{dx}=g'(x)$ . Then $du=g'(x)\ dx$ . Make sure your integral has the form $\int f'(u)\ du$ . Then take the anti-derivative.

Example:

$\int (x^{2}+1)^{3}x\ dx$

Solution:

$u=x^{2}+1,\ du=2x\ dx\ \Rightarrow\ Pick\ u\ and\ differentiate\ to\ find\ du.$

$\int u^{3}\cdot x\ dx\ \Rightarrow\ Substitute\ x^{2}+1=u.$

$\frac{1}{2}\int u^{3}\cdot 2x\ dx\ \Rightarrow\ Pull\ out\ a\ constant\ of\ \frac{1}{2}\ to\ get\ 2x\ dx.$

$\frac{1}{2}\int u^{3}\ du\ \Rightarrow\ Substitute\ du=2x.$

$\frac{1}{2}(\frac{u^{4}}{3})+c\ \Rightarrow\ Take\ the\ integral\ of\ u^{3}.\ using\ the\ power\ rule\ in\ reverse.\ Add\ c.$

$\frac{u^4}{6}+c\ \Rightarrow\ Simplify.$

$\frac{(x^{2}+1)^4}{6}+c\ \Rightarrow\ Substitute\ u=x^{2}+1.\ This\ is\ your\ answer.$

Integration by Parts:

When integrating the product of two function, $f(x)\cdot g(x)$ , you may use integration by parts. This is reverse version of the product rule. Let $u=f(x)$ , and $v=g(x)$ .

$\int u\ dv=uv-\int v\ du$

Often, it does matter which function you choose as $u$ and which is $dv$ . The goal of integration by parts is to reduce the integral $\int u\ dv$ to another integral, $\int v\ du$ , which is simpler to evaluate. We don’t want to make things more complicated.

Tips for choosing $u$ and $dv$ :

- The function which is easiest to integrate, or which simplifies the most when integrated should be chosen as $dv$ .
- The function chosen as $u$ should simplify when differentiated, and should not be too difficult to differentiate.
- If the integration problem becomes more complicated when you apply integration by parts, try picking $u$ and $dv$ another way.

Example:

$\int x^{2}ln(x)\ dx$

Solution:

$ln(x)$ is hard to integrate, but easy to differentiate, so let $u=ln(x),\ and\ dv=x^{2}$ .

$dv=x^{2} \rightarrow v=\int x^{2}\ dx=\frac{x^{3}}{3}\ \Rightarrow\ Integrate\ to\ find\ v.$

$u=ln(x)\rightarrow du=\frac{1}{x}dx\\Rightarrow\ Differentiate\ to\ find\ du.$

$ln(x)(\frac{x^3}{3})-\int\frac{x^{3}}{3}(\frac{1}{x})\ dx\\Rightarrow\ Plug\ these\ into\ the\ formula\ uv-\int v\ du$

$ln(x)(\frac{x^3}{3})-\frac{1}{3}\int x^{2}\ dx\\Rightarrow\ Simplify.$

$ln(x)(\frac{x^3}{3})-\frac{1}{9}x^{3}+C\ \Rightarrow\ Integrate\ and\ simplify.$

Trigonometric Substitution:

Recall the Pythagorean Theorem and the identities of right-angle trigonometry.

Trigonometric substitution allows us to simplify radicals in the integrand of the forms $\sqrt{a^{2}-u^{2}},\ \sqrt{a^{2}+u^{2}},\ and \sqrt{u^{2}-a^{2}}$ to trigonometric functions.

Using the reference triangles above, radicals in the integrand my be replaced as follows:

$\sqrt{a^{2}-u^{2}}=acos(\theta)$
$\sqrt{a^{2}+u^{2}}=asec(\theta)$
$\sqrt{u^{2}-a^{2}}=\pm atan(\theta)\ \Rightarrow$ (positive if u > a and negative if u < -aI)

Example:

$\int \frac{dx}{x^{2}\sqrt{9-x^{2}}}$

Reference triangle:

Solution:

$\sqrt{9-x^2}\rightarrow\ a=3\ \Rightarrow\ Draw\ the\ reference\ triangle\ and\ find\ the\ value\ of\ a$

$x=3sin(\theta)\ \Rightarrow\ Find\ the\ value\ of\ x\ in\ the\ reference\ triangle.$

$dx=3cos(\theta)\ d\theta,\ x^{2}=9(sin(\theta))^{2}\ \Rightarrow\ Find\ the\ values\ of\ dx\ and\ x^2$

$\int\frac{3cos(\theta)\ d\theta}{(9(sin(\theta)^{2})(3cos(\theta))}\ \Rightarrow\ Substitute\ the\ trig\ functions\ into\ the\ integral.$

$\frac{1}{9}\int\frac{d\theta}{(sin(\theta))^{2}}=\frac{1}{9}\int (csc(\theta))^{2}\ d\theta\ \Rightarrow\ Simplify.$

$-\frac{1}{9}cot(\theta)+C\ \Rightarrow\ Integrate.$

$-\frac{1}{9}(\frac{\sqrt{9-x^2}}{x})+C\rightarrow -\frac{\sqrt{9-x^2}}{9x}+C\ \Rightarrow\ Substitute\ for\ cot(\theta)\ using\ reference\ triangle.$

Integration Shortcuts:

Just like rules of differentiation, there are rules you can follow to quickly integrate most functions. here is a list of the rules you should familiarize yourself with. Let C, k, and n be constants.

$\int 9\ dx=C$