Understanding Calculus

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  Table of Contents

  1. Why Study
  2. Numbers
  3. Functions
  4. The Derivative
  5. Differentiation
  6. Applications
  7. Free Falling
  8. Understanding
  9. Derivative
  10. Integration
  11. Understanding
  12. Differentials

  Inverse Functions
  Applications of
  Sine and Cosine
  Sine Function
  Sine Function -
  Differentiation and
  Oscillatory Motion
  Mean Value
  Taylor Series
  More Taylor Series


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Chapter 9 - The Derivative and its Approximations

In the previous section we studied how we could generally approximate the graph of f(x) by studying only its first and second derivative. Now we will develop further on the definition of the derivative to understand how we can obtain accurate approximations of f(x) using its first derivative only. Before continuing let me address a suspicion that is growing in your mind. You are probably wondering why do we care about using functions derivatives to determine the behavior of the original function? If we want to analyze how f(x) behaves we can go and graph it directly, without having to look at its derivative.

To answer this, recall that Calculus was defined as the study of mathematically defined change. In science, changing situations are defined in terms of several conditions or dimensions where one or more dimensions is changing with respect to another dimensions. We need Calculus to be able to define an instantaneous rate of change for the situation. In such a case we need to work backwards from the derivative that we defined to its anti-derivative. The anti-derivative of the function will give us the exact relationship between the situation and the changing dimensions. In order to understand how to define an instantaneous rate of change and then calculate the net change, the relationship between the function and its derivative must be fully grasped. Therefore this chapter will take a look at the definition of the derivative to see how it defines f(x).

Let us begin with the definition of the derivative;

This gives us the exact change in over the infinitely small interval dx. Remember how the instantaneous rate of change was defined as taking the limit as a discrete value for a change in x, goes to zero.

The derivative gives us the instantaneous rate of change of a function over an infinitely small interval, . If we multiply both sides by dx, we get:

If we replace df and dx with and in the above equation, then we get an equation that allows us to approximate the change in a function, . Since the derivative is only defined over an infinitely small interval, we can not use its value to give us the exact change in a function over an interval This is because the derivative of a function changes after each interval, dx and is constant over the interval Therefore the following equation only represents an approximation to the net change in over a discrete interval as it assumes that the rate of change is constant.

Look at the following graph and notice how the approximation gets less and less accurate as Δx increases.

from x=4 to x=6, = 2. The change in f(x) over this interval is by definition,

Substituting the know values in:

approximate this change in by evaluating the rate of change at x=4:

Replacing the infinitesimal values with discrete changes, and , and multiplying both sides by , yields:

Using the derivative we got an error of 4 or 20 %. As Δx get smaller, the error gets less such that this method becomes more accurate as

We can state our observations in a short theorem:

As long as we know a functions value at some initial point we can calculate its value at a nearby point x, by evaluating the derivative at x = a, then multiplying it by Δx=(x-a) and adding the Δf to f(a). This only gives an approximate answer because it assumes the rate of change of the function is constant over the interval Δx.

For example to approximate we can use for f(a). The derivative at this point is

From the calculator we get = 9.1104 which corresponds to an error of less than .01%.

All that we have studied about the derivative until now suggests that the entire graph of a function can be drawn exactly by using the derivative only. This should be of no surprise since the derivative of a function is derived from the function itself, it only remains to undo what is done in the derivation to get the original f(x) back, given only f'(x) The accuracy of our results depends on how small our interval Δx is. This is because, with the exception of linear graphs, the derivative or rate of change of a function is different at each and every point on f(x).

The change in the function is only valid for the derivative evaluated at a point multiplied by an infinitely small dx The derivative is only constant over an infinitely small interval, . The only precise way of defining f(x) in terms of f'(x) is by evaluating f'(x) Δx over infinitely small intervals, keeping in mind that f.(x) is different over each interval. Therefore if we multiply the derivative by Δx we get:

The change in f is exactly equal to the derivative evaluated at a point x, multiplied by an infinitesimal dx. To get accurate approximations of f(x) we need to divide the interval into smaller sub-intervals and evaluate the rate of change at each point. For example if we were to divide the interval from x=4 to x=6 of into 3 intervals to calculate the change in the function.s value from x=4 to x=6 we would have:

Direct calculations tell us that from x=4 to x=6 , the change in f(x) is:

We now want to find this answer using the derivative only. As opposed to our last example we have divided the interval into three sub-intervals where the rate of change is different at each of the three points. In each subinterval Δx = 2/3 =0 .67.

From x=4 to x=4.67 the approximate change in f over this interval is:

From x=4.67 to x=5.33 the approximate change in f over this interval is:

Similarly for the last interval the approximate change in f is:

Therefore the net change in f(x) over the interval from x=4 to x=6 is the sum of the three approximate change in f.s that we calculated. Before stating the results let me first explain some of the notation about to be used. The first sign is:

This tells us to evaluate the change in a function, from two points endpoints, x = a to x = b. The second sign is a summation sum:

The summation signs tells us to add up all the individual values of x from . Returning back to original example we can state our results as, the change in the from x=4 to x=6 is approximately equal to the sum of the calculated for each interval. Mathematically this becomes:

The summation sign can also be written as:

The result corresponds to an error of just 6.6 %, far better than the 20% error we got when we used only one interval of Δx = 2. The reason we still have an error is due to the fact that the derivative is constantly changing over the discrete interval,

We only calculated the derivative at three point from x=4 to x=6. The accurate way to find the precise change in the function would be to let Δx go to zero and then calculate the infinite sum of the along the interval from x=4 to x=6. This is true because according to the definition of the derivative, the rate of change of a function at any point is based on going to zero:

Thus the derivative is defined to be constant only over an infinitely small interval, dx. We shall study this more in depth in the next chapter. Until now we can state the following conclusion:

As n gets larger, ~= 0 and our approximations become exact.

Next section -> Theory of Integration


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