Difference between revisions of "Derivative/Definition"
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− | The [[derivative]] of a [[function]] is defined as the instantaneous rate of change of the function at a certain [[point]]. For a [[line]], this is just the [[slope]]. For more complex [[curves]], we can find the rate of change between two points on the curve easily since we can draw a line through them. | + | The '''[[derivative]]''' of a [[function]] is defined as the instantaneous rate of change of the function at a certain [[point]]. For a [[line]], this is just the [[slope]]. For more complex [[curves]], we can find the rate of change between two points on the curve easily since we can draw a line through them. |
<center>[[Image:derivative1.PNG]]</center> | <center>[[Image:derivative1.PNG]]</center> | ||
− | In the image above, the rate of change between the two points is the slope of the line that goes through them: <math>\frac{f(x+h)-f(x)}h</math>. | + | In the image above, the average rate of change between the two points is the slope of the line that goes through them: <math>\frac{f(x+h)-f(x)}h</math>. |
− | We can move the second point closer to the first one to find a more accurate value of the derivative. Thus, taking the limit as <math>h</math> goes to 0 will give us the derivative of the function at <math>x</math>: | + | We can move the second point closer to the first one to find a more accurate value of the derivative. Thus, taking the [[limit]] as <math>h</math> goes to 0 will give us the derivative of the function at <math>x</math>: |
<center>[[Image:derivative2.PNG]]</center> | <center>[[Image:derivative2.PNG]]</center> | ||
+ | |||
+ | <center><math> f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h. </math></center> | ||
+ | |||
+ | If this limit exists, it is the derivative of <math>f</math> at <math>x</math>. If it does not exist, we say that <math>f</math> is not differentiable at <math>x</math>. This limit is called '''Fermat's difference quotient'''. | ||
+ | |||
+ | == Examples == | ||
+ | We can apply the Fermat's difference quotient to a polynomial of the form <math>f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots + z=0</math> in order to find its derivative. If we imagine the secant line intersecting a curve at the points <math>A</math> and <math>B</math>. Then we can change this to the tangent by setting <math>B</math> on top of <math>A</math>. Let us call the horizontal or vertical distance as <math>h</math>. | ||
+ | |||
+ | <math>\lim_{h\to0} \frac{f(x+h)-f(x)}{h}</math> | ||
+ | |||
+ | <math>\implies \lim_{h\to0} \frac{a(x+h)^n+b(x+h)^{n-1}+c(x+h)^{n-2}+ \cdots z-(ax^n+bx^{n-1}+cx^{n-2}+ \cdots z)}{h}</math> | ||
+ | |||
+ | After canceling like terms we should have all terms contain an <math>h</math>. We can then cancel out the <math>h</math> and set <math>h=0</math>. Our end result is the first-derivative. | ||
+ | |||
+ | The first derivative is denoted as <math>f'(x)</math>. | ||
+ | |||
+ | |||
+ | |||
+ | This would be some tedious work so instead there is a much nicer way to find the derivative. | ||
+ | |||
+ | Let <math>f(x)=3x^n</math>. Let <math>g(x)=x^t+x^{n-1}+5x^{3}</math> | ||
+ | |||
+ | 1. Find <math>f'(x)</math>. | ||
+ | |||
+ | Any function like this is: | ||
+ | <math>f'(x)=3 \cdot n \cdot x^{n-1}=3n \cdot x^{n-1}</math> | ||
+ | |||
+ | |||
+ | |||
+ | 2. Find <math>g'(x)</math>. | ||
+ | |||
+ | Breaking apart on what we used above. | ||
+ | |||
+ | <math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2</math> | ||
+ | |||
+ | <math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2</math> | ||
+ | |||
+ | |||
+ | Let <math>f(x)=-147</math>. Find <math>f'(x)</math>. | ||
+ | |||
+ | |||
+ | If the function <math>f(x)</math> is a constant then its derivative will always be <math>0</math>. | ||
+ | |||
+ | |||
+ | Notation: <math>f'(x)</math> denotes the first derivative for <math>f(x)</math>. The symbol for the second derivative is just <math>f''(x)</math>. For the third derivative it is just <math>f'''(x)</math>. Derivatives are also written as <math>\frac{d}{dx} f(x)</math>. Or if for the nth derivative they are written as <math>\frac{d^n}{dx^n} f(x)</math>. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph. | ||
+ | |||
+ | If <math>f'(x)=6x^2-24</math>. Then the maximum and the minimum occur when: | ||
+ | |||
+ | <math>6x^2-24=</math>, <math>x=2</math> or <math>x=-2</math>. We can plug each back in to the original <math>f(x)</math> if it was given, and the one with the higher y-coordinate is the maximum, while the smaller y-coordinate gives the minimum. | ||
+ | |||
+ | |||
+ | |||
+ | Below are problems for Part I. In Part II(see link below) we will begin to actually "start" the calculus with this. | ||
+ | |||
+ | |||
+ | |||
+ | === Problems === | ||
+ | |||
+ | <math>\boxed{\text{Problem 1}}</math>: Find the first derivative of <math>f(x)</math>, where <math>f(x)=2x^2-15x+7</math>. | ||
+ | |||
+ | |||
+ | <math>\boxed{\text{Solution 1}}</math>: | ||
+ | |||
+ | <math>f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0</math> | ||
+ | |||
+ | <math>f'(x)=4x-15</math>. | ||
+ | |||
+ | |||
+ | |||
+ | <math>\boxed{\text{Problem 2}}</math>: Find the equation of the line tangent to the function <math>f(x)=3x^3-5x^2+12</math> at the point <math>(-1, 4)</math>. | ||
+ | |||
+ | |||
+ | <math>\boxed{\text{Solution 2}}</math>: | ||
+ | |||
+ | We will take the first derivative to determine the slope of the tangent line. | ||
+ | |||
+ | <math>f'(x)=9x^2-10x</math>. If this is the slope of the tangent point then we can just plug <math>-1</math> into the <math>x</math> coordinate to find the actual slope. | ||
+ | |||
+ | <math>f'(x)=9+10=19</math>. The slope of the line is <math>19</math>. | ||
+ | |||
+ | Let the equation be: | ||
+ | |||
+ | <math>y=19x+b</math>. | ||
+ | |||
+ | Plugging <math>(-1,4)</math> in gives <math>4=-19+b</math> and so <math>b=23</math>. | ||
+ | |||
+ | Thus, the equation of the line is <math>y=19x+23</math>. Alternatively, one could use [[point-slope form]] for the line; after determining that the slope is <math>19</math>, as above, this allows one to immediately write down the equation <math>y - 4 = 19(x + 1)</math> of the line. | ||
+ | |||
+ | (Notice that it is implicit in the question that the point <math>(-1, 4)</math> lies on the graph of <math>y = f(x)</math>; it's easy to check that this is actually the case.) | ||
+ | |||
+ | |||
+ | <math>\boxed{\text{Problem 3}}</math>: Find the nth derivative of <math>f(x)=x^n</math> | ||
+ | |||
+ | |||
+ | <math>\boxed{\text{Solution 3}}</math>: | ||
+ | |||
+ | |||
+ | <math>\frac{d}{dx} f(x)=nx^{n-1}</math> | ||
+ | |||
+ | |||
+ | <math>\frac{d^2}{dx^2} f(x)=n(n-1) x^{n-2}</math> | ||
+ | |||
+ | |||
+ | <math>\vdots</math> | ||
+ | |||
+ | |||
+ | <math>\frac{d^{n}}{dx^{n}} f(x)=n(n-1)(n-2) \cdots 1</math> | ||
+ | |||
+ | |||
+ | <math>\frac{d^{n}}{dx^{n}} f(x)=n!</math> | ||
− | |||
− | + | <math>\therefore</math> The nth derivative of <math>f(x)</math> is <math>n!</math>. | |
== See also == | == See also == | ||
* [[Calculus]] | * [[Calculus]] | ||
* [[Derivative]] | * [[Derivative]] | ||
+ | |||
+ | [[Category:Calculus]] |
Latest revision as of 16:21, 3 January 2012
The derivative of a function is defined as the instantaneous rate of change of the function at a certain point. For a line, this is just the slope. For more complex curves, we can find the rate of change between two points on the curve easily since we can draw a line through them.
In the image above, the average rate of change between the two points is the slope of the line that goes through them: .
We can move the second point closer to the first one to find a more accurate value of the derivative. Thus, taking the limit as goes to 0 will give us the derivative of the function at
:
![$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h.$](http://latex.artofproblemsolving.com/8/6/b/86bd376049080d6bd506d0e5955b97b874cfc33f.png)
If this limit exists, it is the derivative of at
. If it does not exist, we say that
is not differentiable at
. This limit is called Fermat's difference quotient.
Examples
We can apply the Fermat's difference quotient to a polynomial of the form in order to find its derivative. If we imagine the secant line intersecting a curve at the points
and
. Then we can change this to the tangent by setting
on top of
. Let us call the horizontal or vertical distance as
.
After canceling like terms we should have all terms contain an . We can then cancel out the
and set
. Our end result is the first-derivative.
The first derivative is denoted as .
This would be some tedious work so instead there is a much nicer way to find the derivative.
Let . Let
1. Find .
Any function like this is:
2. Find .
Breaking apart on what we used above.
Let . Find
.
If the function is a constant then its derivative will always be
.
Notation: denotes the first derivative for
. The symbol for the second derivative is just
. For the third derivative it is just
. Derivatives are also written as
. Or if for the nth derivative they are written as
.
Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.
If . Then the maximum and the minimum occur when:
,
or
. We can plug each back in to the original
if it was given, and the one with the higher y-coordinate is the maximum, while the smaller y-coordinate gives the minimum.
Below are problems for Part I. In Part II(see link below) we will begin to actually "start" the calculus with this.
Problems
: Find the first derivative of
, where
.
:
.
: Find the equation of the line tangent to the function
at the point
.
:
We will take the first derivative to determine the slope of the tangent line.
. If this is the slope of the tangent point then we can just plug
into the
coordinate to find the actual slope.
. The slope of the line is
.
Let the equation be:
.
Plugging in gives
and so
.
Thus, the equation of the line is . Alternatively, one could use point-slope form for the line; after determining that the slope is
, as above, this allows one to immediately write down the equation
of the line.
(Notice that it is implicit in the question that the point lies on the graph of
; it's easy to check that this is actually the case.)
: Find the nth derivative of
:
The nth derivative of
is
.