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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.
 
  
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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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== Differential Calculus Part I ==
  
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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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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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<center><math> f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h. </math></center>
 
  
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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>.
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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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== See also ==
 
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* [[Calculus]]
 
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* [[Derivative]]
 
  
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[[Category:Calculus]]
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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.
 +
 
 +
<math>\boxed{\text{Solution:}}</math>
 +
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 for Part I ==
 +
 
 +
<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 <math>(-1,14)</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>(-1,14)</math> in gives:
 +
 
 +
<math>14=-19+b</math>
 +
 
 +
<math>\implies b=30</math>.
 +
 
 +
 
 +
 
 +
<math>\therefore</math> The equation of the line is <math>y=19x+33</math>.
 +
 
 +
 
 +
 
 +
<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>
 +
 
 +
 
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<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>
 +
 
 +
 
 +
 
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<math>\therefore</math> The nth derivate of <math>f(x)</math> is <math>n!</math>.

Revision as of 19:27, 3 March 2010


Differential Calculus Part I

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.


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.


Long method for Derivative: Let the function be $f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots z=0$. Find the First Derivative.

$\boxed{\text{Solution:}}$ 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 for Part I

$\boxed{\text{Problem 1}}$: 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 $(-1,14)$.


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

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 $(-1,14)$ in gives:

$14=-19+b$

$\implies b=30$.


$\therefore$ The equation of the line is $y=19x+33$.


$\boxed{\text{Problem 3}}$: 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 derivate of $f(x)$ is $n!$.

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