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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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| − | == Differential Calculus Part I ==
| + | 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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| | + | 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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| − | 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.
| + | <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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| − | 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.
| + | 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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| | + | == See also == |
| | + | * [[Calculus]] |
| | + | * [[Derivative]] |
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| − | | + | [[Category:Calculus]] |
| − | 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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| − | <math>\boxed{\text{Solution:}}</math>
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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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| − | <math>\lim_{h\to0} \frac{f(x+h)-f(x)}{h}</math>
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| − | <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>
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| − | 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.
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| − | The first derivative is denoted as <math>f'(x)</math>.
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| − | This would be some tedious work so instead there is a much nicer way to find the derivative.
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| − | 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 <math>f'(x)</math>.
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| − | 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>
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| − | 2. Find <math>g'(x)</math>.
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| − | 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}+15x^2</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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| − | 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.
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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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| − | Below are problems for Part I. In Part II(see link below) we will begin to actually "start" the calculus with this.
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| − | == Problems for Part I ==
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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>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>\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{Solution 2}}</math>:
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| − | 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)=9+10=19</math>. The slope of the line is <math>19</math>.
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| − | Let the equation be:
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| − | <math>y=19x+b</math>.
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| − | Plugging <math>(-1,14)</math> in gives:
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| − | <math>14=-19+b</math>
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| − | <math>\implies b=30</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{Problem 3}}</math>: Find the nth derivative of <math>f(x)=x^n</math>
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| − | <math>\boxed{\text{Solution 3}}</math>:
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| − | <math>\frac{d}{dx} f(x)=nx^{n-1}</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>\vdots</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>\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>.
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goes to 0 will give us the derivative of the function at 
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