Calculus/Differentiation

Prerequisites

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In order to understand and calculate derivatives, one must understand the following topics:

Calculus/Limits

Geometry

Introduction to Differentiation

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Differentiation is the process of finding the unique slope of a line at any point on that line, should such a slope exist. As we shall show, there are commonplace functions that fail this requirement at some point.

Generally, a slope can be found by taking the change in y-coordinates divided by the change in x-coordinates or:

 

where m is the slope.

Now what happens when we attempt to take the slope of a point on a line, or how fast the y's are changing in respect to the x's at a single point? The formula fails when   or  

In order to do this, we have to undergo differentiation.

Definition of a Derivative

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The following formula will calculate the slope of a line at any point:

 

where   indicates the derivative, and h is the difference between a point h units away from x and x



Given  , to find the derivative of f(a) (where a is any x coordinate within the domain of f(x)), use the definition of derivative.

 



Here's an exercise to try:

Find  , given that  . The solution is below.





 


Therefore, the slope of 4x^2 at x=2 is 16.



This is a somewhat tedious process when bigger functions are involved. Take for example:

 


Using the definition of derivative, your equation looks like this:

 


Have fun solving that algebraically!


Luckily, this is where rules for derivatives come in.

Where Derivation Fails

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Two examples of commonplace functions either have ambiguous derivatives or none at some point. The absolute value function has an ambiguous derivative at x=0 For |x|, |x| = -x and the derivative of |x| is -1 if x < 0. |x| = x and the derivative of |x| is 1 if x > 0. But if x=0, then |x|, x, and -x are all zero. No matter how close a number is to zero, the derivative of |x| is 1 for x > 0 but -1 for any x less than zero. Because one cannot determine whether the derivative of |x| is -1, 1, or something else at x=0, one has an ambiguous derivative at x=0... and thus none.

Another function that most off us know well is 1/x. For f(x) = 1/x

  implies


 

 

 

 

 

 

 

No derivative of 1/x can exist at x=0.

Derivative Rules

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In the following table, f(x) and g(x) are functions whereas "a" is a constant number. f'(x) and g'(c) are the derivatives of the functions f(x) and g(x), respectively.

Rule Name Original Function Derivative
Chain Rule f(g(x)) f'(g(x))•g'(x)
constant function derivative a 0
Addition Rule f(x)+g(x) f'(x)+g'(x)
??? x 1
Scalar Rule a•f(x) a•f'(x)
Power Rule (f(x))a a•(f(x))a-1ˆ•f'(x)
Product Rule f(x)•g(x) f(x)•g'(x) + g(x)•f'(x)
Quotient Rule f(x)/g(x) (g(x)•f'(x) - f(x)•g'(x)) / (g(x))2
derivative of trigonometric function sin(x) cos(x)
derivative of trigonometric function cos(x) -sin(x)
derivative of trigonometric function tan(x) sec2(x)
derivative of trigonometric function ln(x) 1/x
derivative of trigonometric function ex ex