## Introduction to Derivatives

Derivatives give us the possibility to easily determine the slope of an equation at any given point by using specific derivative laws. Below, there will be links to videos explaining and demonstrating each of these 4 laws. AS well, we will discuss how to find the complete equation of a tangent line.

## Finding a Derivative / Power Rule

To determine the derivative of a simple monomial, you simply multiply the coefficient by the degree of power from the variable, and then suntact one from the power of the variable. Here is a video working through this situation.

## Sum and Difference Rule

The sum and difference rule is simply a continuation of the power rule. In situations where there are multiple terms in an equation all added or subtracted from one another, simply perform a differentiation on all the separate pieces. There is no need to change signs.

d [ u ± v ] d x = u ′ ± v ′ {\displaystyle {\frac {d[u\pm v]}{dx}}=u'\pm v'}

Sample Question:

Find the derivative of
f
(
x
)
=
3
x
2
+
4
x
4
{\displaystyle \ f(x)=3x^{2}+4x^{4}}

f ′ ( x ) = 6 x + 16 x 3 {\displaystyle \ f'(x)=6x+16x^{3}}

## Product Rule

Here is a video from Khan Academy demonstrating and applying the product rule.

## Quotient Rule

This first video from Khan Academy will demonstrate how the Quotient Rule is derived and how it functions.

This second video will be an example using the Quotient Rule. This video uses the derivatives of tangent sine and cosine which we will learn later, but the method is the same regardless.

## Chain Rule

These videos by Khan Academy will demonstrate the Chain rule. One good way to think about the chain rule is to imagine peeling away an onion layer by layer.

**Introduction Video**

** Definition and Example Video **

** Two Harder Examples **

## Finding the Equation of a Tangent

Finding the equation of a tangent is one of the first possible application of derivatives. Below we will explain the steps to complete this kind of problem, work out a similar problem and also link a video description. There will also be practice questions.

### Steps

1. First take the derivative of the given equation.

2. Use the given co-ordinates in the problem to plug into the newly derived equation in order to find the slope of the tangent line at that specific point.

3. We can now realize that a tangent line will be a straight line so we can use the **y=mx + b** format

4. The slope that we have discovered using the differentiated equation can be **substituted** from **m**

5. Now, we must use the given x and y co-ordinates (is y or x is not not given, use the original equation to determine their value) and substitute them into the x and y positions of our straight line equation.

6. Finally, **solve for b** and we have the straight line equation

Once the sloped is achieved, we may use another form to find the final equation of the tangent.

1. Set the slope equal to (y-(y value given))/(x-(x value given))

2. Solve for y by cross multiplying and other algebraic procedures.

### Video

### Worked out Problem

Find the equation of the tangent line of a slope that is parallel to y=3x^4+5x at (1,0).

1. First take the derivative of the given equation.

y
′
(
x
)
=
12
x
3
+
5
{\displaystyle \ y'(x)=12x^{3}+5}

2. Use the given co-ordinates in the problem to plug into the newly derived equation in order to find the slope of the tangent line at that specific point.

at
(
1
,
0
)
{\displaystyle \ (1,0)}

y
′
(
1
)
=
12
(
1
)
+
5
=
17
{\displaystyle \ y'(1)=12(1)+5=17}

3. We can now realize that a tangent line will be a straight line so we can use the y=mx + b format

y
−
y
o
=
m
(
x
−
x
o
)
+
b
{\displaystyle \ y-y_{o}=m(x-x_{o})+b}

4. The slope that we have discovered using the differentiated equation can be substituted from m

y
−
y
o
=
17
(
x
−
x
o
)
+
b
{\displaystyle \ y-y_{o}=17(x-x_{o})+b}

5. Now, we must use the given x and y co-ordinates (is y or x is not not given, use the original equation to determine their value) and substitute them into the x and y positions of our straight line equation.

y
−
0
=
17
(
x
−
1
)
+
b
{\displaystyle \ y-0=17(x-1)+b}

y
=
17
x
−
17
+
b
{\displaystyle \ y=17x-17+b}

6. Finally, solve for b and we have the straight line equation

0
=
17
(
1
)
−
17
+
b
{\displaystyle \ 0=17(1)-17+b}

b
=
0
{\displaystyle \ b=0}

Solution:
y
=
17
x
−
17
{\displaystyle \ y=17x-17}