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Difference between revisions of "Derivative/Definition"
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Differential Calculus is a sub-field of Calculus that primarily focuses on how functions change as the input changes. In Differential Calculus we usually use Differentiation, or the process of finding the derivative.  
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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.
  
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<center>[[Image:derivative1.PNG]]</center>
  
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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>.
  
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Derivative represents the slope of the slope of the line tangent to a function at some point. We can also find critical points with the first and second derivative.
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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>:
  
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<center>[[Image:derivative2.PNG]]</center>
  
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<center><math> f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h. </math></center>
  
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Long method for Derivative: Let the function be <math>f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots z=0</math>. Find the First Derivative.
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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'''.
  
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<math>\boxed{\text{Solution:}}</math>  
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== Examples ==
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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>.
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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>\lim_{h\to0} \frac{f(x+h)-f(x)}{h}</math>
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Let <math>f(x)=3x^n</math>. Let <math>g(x)=x^t+x^{n-1}+5x^{3}</math>
 
Let <math>f(x)=3x^n</math>. Let <math>g(x)=x^t+x^{n-1}+5x^{3}</math>
  
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1. Find the <math>f'(x)</math>.
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1. Find <math>f'(x)</math>.
  
 
Any function like this is:
 
Any function like this is:
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<math>f'(x)=3 \cdot n \cdot  x^{n-1}=3n \cdot x^{n-1}</math><math>
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<math>f'(x)=3 \cdot n \cdot  x^{n-1}=3n \cdot x^{n-1}</math>
  
  
  
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2. Find </math>g'(x)<math>.
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2. Find <math>g'(x)</math>.
  
 
Breaking apart on what we used above.  
 
Breaking apart on what we used above.  
  
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</math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2<math>
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<math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2</math>
  
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</math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2<math>
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<math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2</math>
  
  
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Let </math>f(x)=-147<math>. Find </math>f'(x)<math>.
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Let <math>f(x)=-147</math>. Find <math>f'(x)</math>.
  
  
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If the function </math>f(x)<math> is a constant then its derivative will always be </math>0<math>.
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If the function <math>f(x)</math> is a constant then its derivative will always be <math>0</math>.
  
  
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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>.
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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>.
  
  
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Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.
 
Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.
  
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If </math>f'(x)=6x^2-24<math>. Then the maximum and the minimum occur when:
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If <math>f'(x)=6x^2-24</math>. Then the maximum and the minimum occur when:
  
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</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.
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<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.
  
  
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== Problems for Part I ==
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=== Problems ===
  
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</math>\boxed{\text{Problem 1}}<math>: Find the first derivative of </math>f(x)<math>, where </math>f(x)=2x^2-15x+7<math>.  
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<math>\boxed{\text{Problem 1}}</math>: Find the first derivative of <math>f(x)</math>, where <math>f(x)=2x^2-15x+7</math>.  
  
  
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</math>\boxed{\text{Solution 1}}<math>:
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<math>\boxed{\text{Solution 1}}</math>:
  
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</math>f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0<math>
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<math>f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0</math>
  
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</math>f'(x)=4x-15<math>.
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<math>f'(x)=4x-15</math>.
  
  
  
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</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 </math>(-1,14)<math>.
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<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>.
  
  
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</math>\boxed{\text{Solution 2}}<math>:
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<math>\boxed{\text{Solution 2}}</math>:
  
 
We will take the first derivative to determine the slope of the tangent line.
 
We will take the first derivative to determine the slope of the tangent line.
  
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</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.
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<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.
  
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</math>f'(x)=9+10=19<math>. The slope of the line is </math>19<math>.
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<math>f'(x)=9+10=19</math>. The slope of the line is <math>19</math>.
  
 
Let the equation be:
 
Let the equation be:
  
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</math>y=19x+b<math>.
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<math>y=19x+b</math>.
  
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Plugging </math>(-1,14)<math> in gives:
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Plugging <math>(-1,4)</math> in gives <math>4=-19+b</math> and so <math>b=23</math>.
  
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</math>14=-19+b<math>
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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.
  
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</math>\implies b=30<math>.
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(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.)
  
  
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<math>\boxed{\text{Problem 3}}</math>: Find the nth derivative of <math>f(x)=x^n</math>
  
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</math>\therefore<math> The equation of the line is </math>y=19x+33<math>.
 
  
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<math>\boxed{\text{Solution 3}}</math>:
  
  
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</math>\boxed{\text{Problem 3}}<math>: Find the nth derivative of </math>f(x)=x^n<math>
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<math>\frac{d}{dx} f(x)=nx^{n-1}</math>
  
  
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</math>\boxed{\text{Solution 3}}<math>:
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<math>\frac{d^2}{dx^2} f(x)=n(n-1) x^{n-2}</math>
  
  
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</math>\frac{d}{dx} f(x)=nx^{n-1}<math>
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<math>\vdots</math>
  
  
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</math>\frac{d^2}{dx^2} f(x)=n(n-1) x^{n-2}<math>
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<math>\frac{d^{n}}{dx^{n}} f(x)=n(n-1)(n-2) \cdots 1</math>
  
  
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</math>\vdots<math>
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<math>\frac{d^{n}}{dx^{n}} f(x)=n!</math>
  
  
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</math>\frac{d^{n}}{dx^{n}} f(x)=n(n-1)(n-2) \cdots 1<math>
 
  
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<math>\therefore</math> The nth derivative of <math>f(x)</math> is <math>n!</math>.
  
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</math>\frac{d^{n}}{dx^{n}} f(x)=n!<math>
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== See also ==
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* [[Calculus]]
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* [[Derivative]]
  
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[[Category:Calculus]]
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</math>\therefore<math> The nth derivate of </math>f(x)<math> is </math>n!$.
 

Latest revision as of 15:21, 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.

Derivative1.PNG

In the image above, the average rate of change between the two points is the slope of the line that goes through them: $\frac{f(x+h)-f(x)}h$.

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 $h$ goes to 0 will give us the derivative of the function at $x$:

Derivative2.PNG
$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h.$

If this limit exists, it is the derivative of $f$ at $x$. If it does not exist, we say that $f$ is not differentiable at $x$. This limit is called Fermat's difference quotient.

Examples

We can apply the Fermat's difference quotient to a polynomial of the form $f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots + z=0$ in order to find its derivative. If we imagine the secant line intersecting a curve at the points $A$ and $B$. Then we can change this to the tangent by setting $B$ on top of $A$. Let us call the horizontal or vertical distance as $h$.

$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

$\implies \lim_{h\to0} \frac{a(x+h)^n+b(x+h)^{n-1}+c(x+h)^{n-2}+ \cdots z-(ax^n+bx^{n-1}+cx^{n-2}+ \cdots z)}{h}$

After canceling like terms we should have all terms contain an $h$. We can then cancel out the $h$ and set $h=0$. Our end result is the first-derivative.

The first derivative is denoted as $f'(x)$.


This would be some tedious work so instead there is a much nicer way to find the derivative.

Let $f(x)=3x^n$. Let $g(x)=x^t+x^{n-1}+5x^{3}$

1. Find $f'(x)$.

Any function like this is: $f'(x)=3 \cdot n \cdot x^{n-1}=3n \cdot x^{n-1}$


2. Find $g'(x)$.

Breaking apart on what we used above.

$g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2$

$g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2$


Let $f(x)=-147$. Find $f'(x)$.


If the function $f(x)$ is a constant then its derivative will always be $0$.


Notation: $f'(x)$ denotes the first derivative for $f(x)$. The symbol for the second derivative is just $f''(x)$. For the third derivative it is just $f'''(x)$. Derivatives are also written as $\frac{d}{dx} f(x)$. Or if for the nth derivative they are written as $\frac{d^n}{dx^n} f(x)$.



Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.

If $f'(x)=6x^2-24$. Then the maximum and the minimum occur when:

$6x^2-24=$, $x=2$ or $x=-2$. We can plug each back in to the original $f(x)$ 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

$\boxed{\text{Problem 1}}$: Find the first derivative of $f(x)$, where $f(x)=2x^2-15x+7$.


$\boxed{\text{Solution 1}}$:

$f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0$

$f'(x)=4x-15$.


$\boxed{\text{Problem 2}}$: Find the equation of the line tangent to the function $f(x)=3x^3-5x^2+12$ at the point $(-1, 4)$.


$\boxed{\text{Solution 2}}$:

We will take the first derivative to determine the slope of the tangent line.

$f'(x)=9x^2-10x$. If this is the slope of the tangent point then we can just plug $-1$ into the $x$ coordinate to find the actual slope.

$f'(x)=9+10=19$. The slope of the line is $19$.

Let the equation be:

$y=19x+b$.

Plugging $(-1,4)$ in gives $4=-19+b$ and so $b=23$.

Thus, the equation of the line is $y=19x+23$. Alternatively, one could use point-slope form for the line; after determining that the slope is $19$, as above, this allows one to immediately write down the equation $y - 4 = 19(x + 1)$ of the line.

(Notice that it is implicit in the question that the point $(-1, 4)$ lies on the graph of $y = f(x)$; it's easy to check that this is actually the case.)


$\boxed{\text{Problem 3}}$: Find the nth derivative of $f(x)=x^n$


$\boxed{\text{Solution 3}}$:


$\frac{d}{dx} f(x)=nx^{n-1}$


$\frac{d^2}{dx^2} f(x)=n(n-1) x^{n-2}$


$\vdots$


$\frac{d^{n}}{dx^{n}} f(x)=n(n-1)(n-2) \cdots 1$


$\frac{d^{n}}{dx^{n}} f(x)=n!$


$\therefore$ The nth derivative of $f(x)$ is $n!$.

See also

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