**Derivatives: Contents** (Click to skip to that section):

## The Basics

## Rules

- The Product Rule
- Chain Rule
- Derivative of functions with exponents (the
**power rule**). - The Quotient Rule

## Specific Functions

- How to Find the Derivative of Simple Functions:
- Derivative of Inverse Functions.
- Derivative of a Trigonometric Function.

## More Advanced Topics

## Common Derivative Rules

This is a list of the more common derivatives (the ones you’ll usually find in the appendix of a text book).

Power of x

c = 0 | x = 1 | x^{n} = n x^{(n-1)} |

Exponential / Logarithmic Derivatives Table

e^{x} = e^{x} |
b^{x} = b^{x} ln(b) |
ln(x) = 1/x |

Trigonometric

sin x = cos x | csc x = -csc x cot x |

cos x = – sin x | sec x = sec x tan x |

tan x = sec^{2} x |
cot x = – csc^{2} x |

Inverse Trigonometric

arcsin x = | 1 / (√ (1- x^{2})) |

arccsc x = | -1 / (|x| √ (x^{2} – 1)) |

arccos x = | -1 / (√ (1- x^{2})) |

arcsec x = | 1 / (|x| √ (x^{2}) – 1) |

arctan x = | 1 / (1 + x^{2}) |

arccot x = | (-1 / 1 +x^{2}) |

Hyperbolic

sinh x = cosh x | csch x = – coth x csch x |

cosh x = sinh x | sech x = – tanh x sech x |

tanh x = 1 – tanh^{2}x |
coth x = 1 – coth^{2}x |

Above is a list of the most common derivatives you’ll find in a derivatives table. If you aren’t finding the derivative you need here, it’s possible that the derivative you are looking for isn’t a generic derivative (i.e. you actually have to figure out the derivative from scratch). If that’s the case and you need to find the derivative, check out the calculus section of this site, or try an online calculator like this one from Wolfram Alpha.

## Product Rule

The **product rule** is used in calculus to differentiate many functions where one function is multiplied by another. The formal definition of the rule is (f * g)’ = f’ * g + f * g’. While this looks tricky, you’re just multiplying the **derivative** of each function by the other function. Recognizing the functions that you can differentiate using the **product rule** in calculus can be tricky. Working through a few examples will aid you in recognizing when to use the product rule and when to use other rules, like the chain rule.

## Product Rule Examples:

## 1. y = x^{3} ln x (Video)

Watch the video or read the steps below:

## How do I Differentiate y=x^{3} ln x?

**The derivative of x ^{3} is 3x^{2}**, but when x

^{3}is multiplied by another function — in this case a natural log, the process gets a little more complicated.

Differentiating functions in calculus that are multiplied with another function is achieved with the **product rule**, a very simple procedure that only involves a little algebra.

Sample problem: Differentiate y = x^{3} ln x.

Step 1: **Name **the first function “f” and the second function “g.” Go in order (i.e. the first listed function should be called “f” and the second should be called “g”).

f = x^{3}

g = ln x

Step 2: **Rewrite the equation** using the new function names f and g you started using in Step 1:

Multiply f by the derivative of g, then add the derivative of f multiplied by g. **You don’t need to actually differentiate at this point: just rewrite the equation.**

y’= x^{3} D (ln x) + D (x^{3}) ln x

Step 3: **Take the derivative of the two functions in the equation you wrote in Step 2. **Leave the two other functions in the sequence alone.

y’= x^{3} (1/x) + (3x^{2} ln x).

Step 4: **Use algebra **to simplify the result. This step is optional, but it keeps things neat and tidy and is almost certainly required by your professor :)

y’= x^{2} + 3x^{2} ln x.

*That’s it!* If you differentiate y=x^{3} ln 3, the answer is y’= x^{2} + 3x^{2} ln x.

## 2. y=(x^{3} + 7x – 7)(5x + 2)

Step 1: Label the first function “f” and the second function “g”.

f = (x^{3} + 7x – 7)

g = (5x + 3)

Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The tick marks mean “derivative” but we’ll use “D” instead.

y’ = (x^{3} + 7x – 7) D(5x + 3) + D(x^{3} + 7x – 7)(5x + 3)

Step 3: **Take the derivative** of the two functions identified in the equation you wrote in Step 2.

y’ = (x^{3} + 7x – 7) (5) + (3x^{2} + 7)(5x + 3)

Step 4: Use algebra to multiply out and neaten up your answer:

y` = 5x^{3} + 35x – 35 + 15x^{3} + 9x^{2} + 35x + 21 = 20x^{3} + 9x^{2} + 70x – 14

That’s it!

## 3. y = x^-3 (17 + 3x^-3)

**Sample problem**: Differentiate y = x^{-3}(17 + 3x^{-3}) using the product rule.

Step 1: Name the functions so that the first function is “f” and the second function is “g.” In this example, we have:

f = x^{-3} and

g = (17 + 3x^{-3})

Step 2: **Rewrite the equation:** Multiply f by the derivative of g, added to the derivative of f multiplied by g.

f’ = x^{-3 }D (17 + 3x^{-3}) + D(x^{-3}) (17 + 3x^{-3}).

Step 3: Take the two **derivatives** of the equation from Step 2:

f’= x^{-3 } (-9x^{-4}) + (-3x^{-4}) (17 + 3x^{-3}).

Step 4: Use **algebra** to expand and simplify the equation:

f’ = -9x^{-7}-51x^{-4}-9x^{-7} = -18x^{-7}-51x^{-4}.

That’s it!

## 4. y = 6x^{3/2} cot x.

Step 1: Label the first function “f” and the second function “g”.

f = 6x^{3/2}

g = cot x

Step 2: Rewrite the functions: multiply the first function f by the derivative of the second function g and then write the derivative of the first function f multiplied by the second function, g. The tick(‘) in the formal definition means “derivative” but we’ll use “D” instead.

y’ = (6x^{3/2})* D (cot x) + D(6x^{3/2})* cot x

Step 3: Take the derivative of the two functions from Step 2.

y’ = (6x^{3/2})* (– csc^{2} x) + (6^{(3/2)}x^{1/2})* cot x

Step 4: Use algebra to multiply out and neaten up your answer:

f’ = 6x^{3/2} – csc^{2} x + 9x^{1/2}cot x

= 3x^{1/2}(2x – csc^{2} x + 3 cot x)

That’s it!

Tip: Don’t be tempted to skip steps, especially when multiplying out algebraically. Although you might think you’re in **calculus** (and therefore know it all when it comes to algebra!), common mistakes usually happen in **differentiation** not by the actual differentiating process itself, but when you try and multiply out “in your head” instead of being careful to multiply out piece-wise.

## The Quotient Rule

## The Quotient Rule

The

**quotient rule**is used to differentiate functions that are being divided. If you have a function f(x) (top function) and g(x) (bottom function) then the quotient rule is:

**[f(x)/g(x)] = [f’(x)g(x) – g’(x)f(x)] / [g(x)] ^{2}. **

It looks ugly, but it’s nothing more complicated than following a few steps (which are **exactly the same** for each quotient). You might also notice that the numerator in the quotient rule is the same as the product rule with one slight different–the addition sign has been replaced with the subtraction sign.

Working through a few examples will help you in recognizing when to use the quotient rule and when to use other rules, like the chain rule.

**Sample Problem** Differentiate the following function:

y = 2 / (x + 1)

**Solution**:

**Note**: I’m using D as shorthand for derivative here instead of writing g'(x) or f'(x):

**Identify g(x) and h(x)**. The top function (2) is g(x) and the bottom function (x + 1) is f(x).**Plug your functions (from Step 1) into the formula:**

y’ = D{2} (x + 1) – D {x + 1} (2) / (x + 1)^{2}**Work out your derivatives**. For example, the derivative of 2 is 0. y’ = (0)(x + 1) – (1)(2) / (x + 1)^{2}**Simplify**: y’ = -2 (x + 1)^{2}

When working with the quotient rule, always start with the bottom function, ending with the bottom function squared.

The following three examples will show you how to work through the Quotient Rule:

## How to Differentiate (2x + 1) / (x – 3)

The **quotient rule** is just the product rule rearranged. For example, 10 * 0.5 = 5 is the same as 10 / 2 = 5. The two different rules are used in calculus for the same reason the two rules are used in basic math: because it’s easier to memorize that way.

## How to Differentiate (2x + 1) / (x – 3): Steps

Step 1: Name the top term, the denominator, f(x) and the bottom term, the numerator, g(x). This gives you two new functions:

f(x) = 2x + 1

g(x) = x – 3

Step 2: **Place your functions** f(x) and g(x) into the quotient rule. We’ll use d/dx here to indicate a derivative.

f'(x) = (x – 3) d/dx [2x + 1] – (2x + 1) d/dx[x – 3] / [x-3]^{2}

Step 3:Differentiate the indicated functions in Step 2. In this example, those functions are [2x + 1] and [x + 3].

f'(x) = (x – 3)(2)-(2x + 1)(1) / (x – 3)^{2}

Step 4:**Use algebra** to simplify where possible. The solution is 7/(x – 3)^{2}.

## How to Differentiate tan(x)

The **quotient rule** can be used to differentiate tan(x), because of a basic quotient identity, taken from trigonometry: tan(x) = sin(x) / cos(x).

Step 1: **Name** the top term f(x) and the bottom term g(x). Using our quotient trigonometric identity tan(x) = sinx(x) / cos(s), then:

f(x) = sin(x)

g(x) = cos(x)

Step 2: **Place your functions** f(x) and g(x) into the **quotient rule**. The term d/dx here indicates a derivative.

f'(x) = cos(x) d/dx[sin(x)] – sin(x) d/dx[cos x]/[cos]^{2}

Step 3:Differentiate the indicated functions from Step 2. In this example, those functions are [sinx(x)] and [cos x].

f'(x)= cos^{2}(x) + sin^{2}(x) / cos^{2}x.

Step 4:**Use algebra** to simplify where possible. The solution is 1/cos^{2}(x), which is equivalent in trigonometry to sec^{2}(x).

## How to Differentiate 2^{x}/2^{x}x

In this sample problem, you’ll need to know the algebraic rule that states: a^{x}a^{x} = a^{x + x} = a^{2x} and a^{x}b^{x} = (ab)^{x}.

Step 1: **Name** the top term f(x) and the bottom term g(x).

f(x) = 2^{x}

g(x) = 2^{x} – 3^{x}.

Step 2: **Place the functions** f(x) and g(x) from Step 1 into the **quotient rule**. The term d/dx here indicates a derivative.

f'(x) = (2^{x} – 3^{x}) d/dx[2^{x}] – (2^{x}) d/dx[2^{x} – 3^{x}]/(2^{x} –

3^{x})^{2}.

Step 3: Differentiate the indicated functions (d/dx)from Step 2. In this example, those functions are 2^{x} and [2^{x} – 3^{x}]

f'(x)= (2^{x} – 3^{x}) d/dx[2^{x} ln 2] – (2^{x})(2^{x}2^{x} ln 2 – 3^{x} ln 3).

Step 4:** Use algebra** to simplify where possible (remembering the rules from the intro).

f'(x) = 2^{2x} ln 2 – 6^{x} ln 2 – (2^{2x} ln 2 – 6^{x} ln 3) / (2^{x} – 3^{x})^{2}

By simplification, this becomes:

f'(x) = 6^{x}(ln 3 – ln 2) / (2^{x}-3^{x})^{2}

## Common Derivatives

## Derivative of a Constant

While most of the rules require you to complete several steps, taking the derivative of a constant only requires you to perform **one step**:

Change the constant to a 0, because the derivative or slope of any constant function is equal to zero.

In other words, if f(x)=c, then f'(x)=0.

**Examples of how to take the Derivative of a Constant:**

If f(x) = 5, then f'(x) = 0

If f(x) = 0.1, then f'(x) = 0

If f(x) = 10000000, then f'(x) = 0.

If f(x) = 55 1/3, then f'(x) = 0

If f(x) = √9, then f'(x) = 0

*That’s it!*

**Warning**: The rule that the derivative of a constant *only* applies if you take the derivative of a constant, and not constants that also have exponents, constants multiplied by x, or anything other than a number. While √9 is a constant, √9x is not. If in doubt, graph your function. If the result is a horizontal line, then your function is a constant.

## Derivative of x

The derivative of x = 1. Similarly, the derivative of -x = 1.

**Why**?

The definition of the derivative is the slope of the tangent line at any point on the graph. The function y=x is a constant function. It has a positive slope of exactly 1 at all points on the graph — that’s why the derivative for the whole function is defined as 1.

The graph of -x is a decreasing graph with a negative slope of exactly -1 at all points:

One you wrap the idea around your head that the derivative is just the slope of the tangent line, it makes finding simple derivatives extremely easy. If only all derivatives in calculus were this simple!

## Derivative of 2x

The derivative of any x value multiplied by a constant is just the constant. For example, the derivative of 2x is 2, or the derivative of 100x is 100. You can apply this rule to *any* x value multiplied by a constant, including pi, e, decimals, fractions and constants like z,p,w, or v.

**Why?**

The derivative is a tangent line at a point. In other words, find the slope at a point and you have the derivative. The slope of the line 2x is 2, no matter what point you pick to find the slope. Therefore, the derivative of the entire function is 2.

**Tip**: Just in case you need a refresher, the slope formula is change in y / change in x. You can use this formula to take an average of a slope at two points; as the slope for a linear graph (like 2x) is constant, finding the slope between two points will also give you the derivative of 2x.

## Derivative of 3x

The derivative of any x value multiplied by a constant is just the constant. For example, the derivative of 99x is 99, or the derivative of 101x is 101.

**Why?**

The derivative is defined as the tangent line at a point. To put it another way, all you need to do is find the slope at a particular point and that value is the derivative.

You can find the slope of a line between two points by using the slope formula:

Slope = Change in y / change in x.

As you can probably deduce from the formula, it’s impossible to find the slope at a **point**…because there’s no change! In calculus, if you want to find the slope at a point you just pick a couple of points that are very close to the point you want to find the slope for. For example, if you want to find the derivative of 3x (which is just the slope), you might pick the point x=3 to find the derivative at. To use the slope formula, you need two points, so you could choose x = 2 and x = 4 (which are 1 units either side of 3). A linear function has a constant slope — so it doesn’t matter which points you pick!

The slope of the line 3x is 3, no matter what point you pick to find the slope. Therefore, the derivative of 3x is the derivative of the entire function: 3.

## Derivative of e

The derivative of e is 0.

**Why? **

*Because the derivative of any constant is 0.
*

“e“, sometimes called Napier’s constant, is not a variable like x or y. It’s a constant like π. It has a value of approximately 2.718. This graph shows the function y = e (red) and y = e

^{x}(green):

If you look at the graph of e, you can see that the slope is zero for all points on the line; A horizontal line always has a slope of zero. Therefore, the derivative is always zero for constant functions (like e) that produce a horizontal line when graphed.

However, it becomes slightly more complicated when you try to find a derivative of e when it’s combined with another function. For example, you might be asked to find the derivative of e functions that look like this: e^{x} or x^{2x2}. For these functions, you need to use a rule called the chain rule. See: How to differentiate e functions using the chain rule.

Natural Log (ln)

A **natural logarithm** of a function, written ln(x) and sometimes log(subscript e)(x), is a logarithm that is equal to the power the natural number, e, would have to be raised to in order to equal x. Given that a logarithm is essentially a function, this adds some unique issues when attempting to differentiate natural logs.

Sample problem # 1: Differentiate the function ln(x).

d/dx ln(x) = 1/x

*That’s all!*

Sample problem # 2: Differentiate the function ln(sqrt(x))

*Note:At a glance, this would simply be 1/sqrt(x), but the reality is a bit
trickier than that.*

Step 1:Use the law for algorithms that states ln x

^{n}= n ln x to separate

the exponent from the function. As sqrt(x) is the same as x

^{1/2}, this puts it in a form we can apply the earlier rule to.

d/dx ln(sqrt(x)) = d/dx ½ ln x

Step 2: Find the derivative of the function. Given that ln x derives to 1/x, and a constant’s derivative is always itself:

d/dx ½ ln x = 1/2x

*Note: d/dx xy is equal to d/dx x d/dx y.*

Tip:Whenever you come across functions you need to differentiate that include natural logs, you can substitute 1/x for the derivative of the

natural log of x. It is important to note that this applies specifically to

the **natural log**, and not to logarithms of any other base.

## The derivative of sin^{3}x

**The derivative of sin ^{3}x is 3sin^{2}x cos x. **

There are two main ways to arrive at the derivative, either by using the definition of a limit (the long way), or by using a shortcut, called the general power rule. Shortcuts exist so that you can skip using the long way of finding a derivative: the definition of a limit. The general form of the power rule helps you to differentiate functions of the form [u(x)]

^{n}], like sin3x, which can be rewritten as [sin x]

^{3}, which has the inside function of “sin x” and the outside function of x

^{3}. The general form of the power rule is:

If y-u

^{n}, then y = nu

^{n – 1}*u’, where “u” is the inside function.

**Sample problem**: Find the Derivative of Sin3x

Step 1: **Rewrite the equation** to make it a power function:

sin^{3}x = [sin x]^{3}

Step 2: **Find the derivative for the “inside” part of the function**, sin x. According to the general rules for differentiation, the derivative of sin x is cos x:

f’ sin x = cos x

Step 3: **Rewrite the function **according to the general power rule. In other words, write out the general power rule, substituting in your function where appropriate. The last half of the general power rule is the derivative of the inside function you worked out in Step 2:

f- = 3[sin x]^{3-1}[cos x] = 3[sin x]^{2}[cos x]

Step 4: **Rewrite using algebra**:

3[sin x]^{2}[cos x] = 3sin^{2}x cos x

*That’s it!*

**Tip: **Calculus uses a lot of algebra and trigonometry. If your algebra skills are weak, this is where the course will likely become difficult. Rather than concentrating on memorizing the rules of differentiation, concentrate on improving your algebra skills. Being able to look at a function and seeing which rule might apply if you manipulate the equation (for example, knowing that a square root can be rewritten as “to the 1/2 power”) is key to working out derivatives.

## How to differentiate exponents

The **Power Rule** is one of the first rules you will come across in differential calculus, and with the prevalence of **exponents **in calculus it is one you will use often. The Power Rule is stated as “The derivative of x to the nth power is equal to n times x to the n minus one power,” when x is a monomial (a one-term expression) and n is a real number. In symbols it looks as follows:

d/dx x^{n} = nx^{n – 1}

The rule is succinct and simple. Place the exponent in front of “x” and then subtract 1 from the exponent. For example, d/dx x^{3} = 3x^{(3 – 1)} = 3x^{2}.

*That’s it!*

## Differentiate Exponents: Steps

With the

**power rule**, you can quickly move through what would be a complex differentiation in seconds without the aid of a calculator. Take the derivative of x

^{1000}for example. Attempting to solve (x+h)

^{1000}would be a time-consuming chore, so here we will use the Power Rule.

Step 1: Find “n”, which is the exponent. For this problem, n is equal to 1000.

Step 2: Substitute the value “n” into the front of the base to get 1000x^{1000}.

Step 3: Subtract 1 from the exponent:

1000x^{1000-1} = 1000x^{999}

That’s it!

## Constant multiplied by a power rule function

The derivative of a constant is always zero and the derivative of a function depends upon what kind of function it is (for example, you can differentiate exponents with the power rule). Intuitively, you might think that a constant multiplied by a function is zero, because the derivative of a constant is zero (0 * anything = 0). However, differentiation in calculus isn’t always intuitive; the derivative of a constant multiplied by a power rule function is actually equal to the constant times the derivative of the function.

**Sample Question 1:** What is the derivative of 5x^{3}?

Step 1:** Separate the constant from the function.**

5

x^{3}

Step 2:** Differentiate the function **using the rules of differentiation. The function x^{3} is an exponent and so is differentiated using the power rule:

d/dx [x^{3}] = 3x^{3 – 1} = 3x^{2}

Step 3: **Place the constant back in front of the derivative** of the function from Step 2:

5(3^{x2})

Step 4:** Use algebra to multiply through:**

5(3^{x2}) = 15x^{2}

**Sample Question 2:** What is the derivative of -7x^{-4}?

Step 1:** Separate the constant from the function.**

-7

x^{-4}

Step 2:** Differentiate the function **using the rules of differentiation. The function x^{-4} is an exponent and so is differentiated using the power rule:

d/dx [x^{-4}] = -4x^{-4-1} = -4x^{-5}

Step 3: **Place the constant back in front of the derivative** of the function from Step 2:

-7[-4x^{-5}]

Step 4:** Use algebra to multiply through:**

-7[-4x^{-5}] = 28x^{-5} = 28/x^{5}

*That’s it!*

## Proof of the Power Rule

In order to understand how the proof of the power rule works, you should be familiar with the binomial theorem (although you might be able to get away with *not *knowing it if your algebra skills are strong). If you need a refresher, see this article on how to use the binomial theorem. You’ll also need to be comfortable with the formal definition of a limit *and *you’ll need strong algebra skills. If you have those three prerequisites, it should be very easy to follow. This proof of the power rule is the proof of the **general form of the power rule**, which is:

In other words, this proof will work for *any* numbers you care to use, as long as they are in the power format.

**Sample problem: **Show a proof of the power rule using the classic definition of the derivative — the limit.

Step 1: **Insert the power rule into the limit definition:**

Step 2: **Use the binomial theorem to evaluate the equation from Step 1:**

Note: I included “…” to indicate this is an incomplete series. In order to prove the power rule you don’t need to write out the entire series.

Step 3: **Simplify the equation** from Step 2 using algebra. Basically, you’re canceling out any +n^{n} and -n^{n}, and dividing by Δx:

Step 4: **Delete the terms that multiply by δx **(because δx is such an insignificant amount it’s practically zero):

Step 5: **Expand the equation**, using combinations (n choose 1):

Step 6: **Use the following rules to further reduce the equation**:

- 1! reduces to one, so you can eliminate it.
- 7!/6! = 7 or 10!/9!= 10, so n!/n-1! = n

This equation is the derivative of X^{n}.

*That’s it! *

## Derivative of Inverse Functions

An inverse function is a function that undoes another function; you can think of a function and its inverse as being opposite of each other. The slopes of inverse linear functions are **multiplicative inverses **of each other. For example, a linear function that has a slope of 4 has an inverse function with a slope of ^{1}⁄_{4}. This means that you can find the derivative of inverse functions by using a little geometry or you could find the derivative of inverse functions by finding the inverse function for the derivative and then using the usual rules of differentiation to differentiate the inverse function. However, there is a third option: using the formula to find the derivative of inverse functions.

**Sample problem:** *Find the derivative of the inverse function for the following function:*

**Part One: Find the inverse of the function**

Step 1: Swap f(x) for y:

Step 2: Switch x and y:

Step 3: Solve for y, using algebra, to get the inverse function:

- x
^{2}= y-3 - y = x
^{2}+ 3

y = x^{2} + 3 is the inverse function

*Note: You could find the derivative of the inverse function at this point, using the usual rules for differentiation. Continue with the Steps if you want to use the formula for the derivative of inverse functions.*

**Part Two: Using the Formula for the Derivative of Inverse Functions**

Step 4: Calculate the derivative for the original function. Use the chain rule for this sample problem.

Step 5: Insert your answer from Step 4 into the derivative of inverse functions formula:

Step 6: Replace the “x” from your answer in Step 5 with the inverse function from Step 3:

Note that the square and square root will cancel, so will the 3s, leaving **2x** as the derivative of the inverse function.

*That’s it!*

**Tip: ** In order for the derivative of the inverse function to work, the inverse function must be differentiable at f^{-1}(x) and f'(f^{-1}(x)) is not equal to zero.

**Warning:** If you flip the graph of this sample function, you only get half of the parabola. Therefore, this particular inverse only holds for x>0.

## How to Find the Derivative of a Trigonometric Function

Trigonometric functions, also called circular functions, are functions of angles. You’re probably already familiar with the six trigonometric functions: sin x, cos x, tan x, sec x, csc x and cot x.

**Derivatives for the six trig functions**:

- d/dx sin x = cos x
- d/dx csc x = -csc x cot x
- d/dx cos x = – sin x
- d/dx sec x = sec x tan x
- d/dx tan x = sec
^{2}x - d/dx cot x = – csc
^{2}x

However, you might be asked (especially in beginning calculus) to find the derivative of a trig function using the definition of a derivative instead of a table. When you use the definition of a derivative, you’re actually working on a proof. In other words, if you want to prove that one function is a derivative of another, you’ll nearly always start with the definition of a derivative and end with the derivative of the trigonometric function.

**Sample Problem**: Find the derivative of a trigonometric function (sin x) using the definition of a derivative (in other words, prove that d/dx sin x = cos x:

Step 1: **Insert the function sin x into the definition of a derivative:**

Step 2: **Use the trigonometric identity **sin(a+b)=sin a * cos B + cos a * sin B to rewrite the definition from Step 1:

Step 3: **Use algebra** to rewrite the formula in Step 2:

= – sin x* (0) + cos x * (1) = cos x

*That’s it!*

Tip: You can use the exact same technique to work out a proof for any trigonometric function. Start with the definition of a derivative and identify the trig functions that fit the bill.

## Find the Derivative Using the Derivative Formula

There are short cuts to finding derivatives (like the ones above), but when you first start learning calculus you’ll be using the derivative formula: f'(x) = lim _{Δx → 0} ( f( x + Δx ) – f (x) ) / Δx.

**Sample problem #1:**Find the derivative of f(x) = √(4x + 1)

Step 1:**Insert the function into the derivative formula.** The function is √(4x + 1), so:

f'(x) = lim _{Δx → 0} √( 4( x + Δx ) + 1 – √(4x + 1) ) / Δx.

If this looks confusing, all we’ve done is changed “x” in the formula to x + Δx in the first part of the derivative formula.

Step 2: **Use algebra to work the formula.** Here’s where you’ll benefit from strong algebra skills, because every formula is different.

*Multiply the top and the bottom*by √( 4( x + Δx ) + 1**+**√(4x + 1):

f'(x) = lim_{Δx → 0}√( 4( x + Δx ) + 1 – √(4x + 1) ) * √( 4( x + Δx ) + 1 + √(4x + 1) / Δx * √( 4( x + Δx ) + 1 + √(4x + 1)

which reduces to:

= lim_{Δx → 0}4(x + Δx) + 1 – (4x + 1) / Δx (√ (4x + Δx) + 1) + √ 4x + 1*Distribute the 4:*

= lim_{Δx → 0}(4x + 4Δx + 1 – 4x – 1) / (Δx (√ (4x + Δx) + 1) + √ (4x + 1)*Delete terms.*In this case you can delete 4x, Δx and 1.

= lim_{Δx → 0}4 / ((√ (4x + Δx) + 1) + √ 4x + 1))

Step 3:*Take the limit.* The Δx will drop out (because it’s an insignificant increment). Again, strong algebra skills will help here:

= 4 / ((√ (4x + 1) + √ 4x + 1)

= 4 / 2 √(4x + 1)

= 2 / √(4x + 1)

*That’s it!*

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## Implicit Differentiation

Implicit differentiation is used when it’s difficult, or impossible to solve an equation for x. For example, the functions y=x^{2}/y or 2xy = 1 can be easily solved for x, while a more complicated function, like 2y

^{2}-cos y = x

^{2}cannot. When you have a function that you can’t solve for x, you can still differentiate using implicit differentiation. With this technique, you directly differentiate both sides of the equation

*without solving for x*.

## Implicit Differentiation

**Sample problem #1:** Differentiate 2x-y = -3 using implicit differentiation.

Step 1: **Write out the function** with the derivative on both sides:

dy/dx [2x-y] = dy/dx [-3]

This step isn’t technically necessary but it will help you keep your calculations tidy and your thoughts in order.

Step 2: **Differentiate the right side of the equation**. The right side of this equation is a constant, so the derivative is zero:

dy/dx [2x-y] = 0

Step 2: **Differentiate the left side of the equation**. The derivative of 2x-y is 2 (using the power rule and constant rule). Remember to treat the dependent variable as a function of the dependent variable:

2- dy/dx = 0

Step 4: **Use algebra to solve for the derivative**.

dy/dx = 2.

*That’s it!*

**Sample problem #2:** Differentiate y^{2} + x^{2} = 7 using implicit differentiation.

Step 1: **Differentiate the left and right sides of the equation**. This example also uses the power rule and constant rule:

2y dy/dx + 2x = 0

Step 2: **Use algebra to solve**:

2y dy/dx + 2x = 0

2y dy/dx = -2x

dy/dx = -2x/2y

dy/dx = -x/y

*That’s it!*

**Tip:** These basic examples show how to perform implicit differentiation using the power rule and constant rule. Depending on what function you are trying to differentiate, you may need to use other techniques of differentiation, including the chain rule, to solve.

Back to Top.

## How to find critical numbers

A critical number is a number “c” that either makes the derivative equal to zero or it makes the derivative undefined. Critical numbers indicate where a change in the graph is taking place — for example, an increasing to decreasing point or a decreasing to increasing point. The number “c” also has to be in the domain of the original function (the one you took the derivative of). Finding critical numbers is an easy task if your algebra skills are strong; unfortunately, if you have weak algebra skills you might have trouble finding critical numbers. Why? Because each function is different, and algebra skills will help you to spot undefined domain possibilities such as division by zero. If your algebra isn’t up to par — now is the time to restudy the old rules!

## How to find critical numbers

**Sample question:** Find the critical numbers for the following function: ^{x2}⁄_{x2-9}

Step 1: **Take the derivative of the function**. Using the quotient rule, we get:

^{-18x}⁄_{(x2 – 9)2}.

Step 2: **Figure out where the derivative equals zero.** This is where a little algebra knowledge comes in handy, as each function is going to be different. For this particular function, the derivative equals zero when -18x = 0 (making the numerator zero), so one critical number for x is 0 (because -18(0) = 0). Another set of critical numbers can be found by setting the denominator equal to zero, you’ll find out where the derivative is undefined:

(x^{2} – 9) = 0

(x – 3)(x + 3) = 0

x = ±3

Step 3: **Plug any critical numbers you found in Step 2 into your original function** to check that they are in the domain of the original function. For this particular function, the critical numbers were 0, -3 and 3.

f(x) = ^{02}⁄_{02-9} = 0. Therefore, 0 is a critical number.

For +3 or -3, if you try to put these into the denominator of the original function, you’ll get **division by zero**, which is **undefined**. That means these numbers are not in the domain of the original function and are not critical numbers.

*That’s it!*

Back to Top.

## First Derivative Test

The **first derivative test** is one way to study increasing and decreasing properties of functions. The test helps you to:

- Find the intervals where a function is decreasing or increasing.
- Identify local minima and maxima.
- Sketch a graph without the aid of a graphing calculator.

**It’s useful to think of the derivative here as just the slope of the graph.** Technically, it’s the *slope of the tangent line at a certain point*, but simplifying the concept to just increasing or decreasing slopes helps with this particular test.

Your result from the first derivative test tells you one of

**three things**about a continuous function:

- If the first derivative (i.e. the slope) changes from
**positive to negative**at a certain point (going from left to right on the number line), then the function has a*local maximum*at that point. Points**b**and**c**on the above graph are examples of a local maximum. - If the first derivative changes from
**negative to positive**(going from left to right on the number line), then the function has a*local minimum*at that point. Point**c**on the graph is a local minimum. - If the first derivative doesn’t change sign at the critical number (going from left to right on the number line), then there is
**neither**a local maximum or a local minimum at that critical number. Point**e**is one example where the slope does not change sign.

*Sample question: **Use the first derivative test to find the local maximum and/or minimum for the graph x ^{2} + 6x + 9 on the interval -5 to -1.*

Step 1:

**Find the critical numbers for the function**. (Click here if you don’t know how to find critical numbers).

- Taking the derivative: f’= 2x + 6
- Setting the derivative to zero: 0 = 2x + 6
- Using algebra to solve: -6 = 2x then -6/2 = x, giving us x =-3

There is one critical number for this particular function, at x=-3.

Step 2: **Choose two values close to the left and right of the critical number**. The critical number in this example is x =-3, so we can check x = -2.99 and x = 3.01 (these are arbitrary, but pretty close to -3; you could try -2.999999 and -3.0000001 if you prefer).

Step 3: **Insert the values you chose in Step 2**, into the derivative formula you found while figuring out the critical numbers in Step 1:

*For x = -3.01:*

f’ = 2(-3.01) + 6 = -0.02 → a negative slope

*For x = -2.99*

f’= 2(-2.99) + 6 = 0.02 → a positive slope

Step 4: **Compare your answers to the three first derivative test rules** (stated in the intro above). The derivative changes from negative to positive around x = -3, so there is a local minimum.

*That’s it!*

**Tip:** To check that you found the correct critical numbers, graph your equation. As the graph above clearly shows — you should only find one critical number for this particular equation, at x =-3.

**The following table summarizes the application of the first derivative test (f’) and the second derivative test (f”) for drawing graphs.**

## Second Derivative Test

The second derivative test is used in calculus to find intervals where a function has a relative maxima and minima. You can also use the second derivative test to determine concavity.

## The second derivative test for extrema

The second derivative test for extrema uses critical numbers to state that:

- If the second derivative f” at a critical value is positive, the function has a relative minimum at that critical value.
- If the second derivative f” at a critical value is negative, the function has a relative maximum at that critical value.
- If the second derivative f” at a critical value is inconclusive the function
*may*be a point of inflection.

## The second derivative test for concavity

The second derivative test for concavity states that:

- If the second derivative is greater than zero, then the graph of the function is concave up.
- If the second derivative is less than zero, then the graph of the function is concave down.

**Sample problem:**What concavity does the graph x^{3}have between -2 and 3? Where are the local minimum(s) and maximum(s)?Step 1: **Find the critical values for the function**. (Click here if you don’t know how to find critical values).

*Take the derivative*: f’= 3x^{2} – 6x + 1.

*Set the derivative equal to zero*: 0 = 3x^{2} – 6x + 1.

*Solve for the critical values (roots), using algebra*.

There are two critical values for this function:

C_{1}:1-^{1}⁄_{3}√6 ≈ 0.18.

C_{2}:1+^{1}⁄_{3}√6 ≈ 1.82.

Step 2: **Take the second derivative** (in other words, take the derivative of the derivative):

f’ = 3x^{2} – 6x + 1

f” = 6x – 6 = 6(x – 1).

Step 3: **Insert both critical values into the second derivative**:

C_{1}: 6(1 – ^{1} ⁄_{3}√6 – 1) ≈ -4.89

C_{2}: 6(1 + ^{1} ⁄_{3}√6 – 1) ≈ 4.89.

The second derivative at C_{1} is negative (-4.89), so according to the second derivative rules there is a local maximum at that point.

The second derivative at C_{1} is positive (4.89), so according to the second derivative rules there is a local minimum at that point.

Step 4: **Use the second derivative test for concavity** to determine where the graph is concave up and where it is concave down. For this function, the graph has negative values for the second derivative to the left of the inflection point, indicating that the graph is concave down. The graph has positive x-values to the right of the inflection point, indicating that the graph is concave up.

The above graph shows x^{3} – 3x^{2} + x-2 (red) and the graph of the second derivative of the graph, f” = 6(x – 1) green. Positive x-values to the right of the inflection point and negative x-values to the left of the inflection point.

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