Derivative

Derivative

The concept of a derivative is usually introduced graphically. A visual perspective can be easier to grasp. I will take that approach as well. We will start with the graph of a line and the meaning of its slope.  Then we will rewrite the equation for the slope of the line in such a way to develope the idea of the derivative. First, review the equations of a line. Consider the following graph of a line. (Important concepts from the graphs are repeated within subsequent text.) The slope at x = x0 is m.  The slope at x = x1 is also m. In fact, the slope at any value of x is m.   The slope m defines how y changes as x changes.  This change is called the rate of change of y with respect to x.    The slope at a given point on this line is called the instantaneous rate of change of y with respect to x.   We will later call this the derivative of the function f at that value of x. For this line, since the slope m is constant at each point, the instantaneous rate of change is also constant. Let's continue with this graph. We can write y1 = f(x1).   Likewise, y0 = f(x0). To facilitate our discussion of a derivative we can also set h to equal the difference between x1 and x0, like so:  h = x1 - x0.   We will now rewrite the denominator simply as h. Using h, we then have x1 = x0 + h.     Substituting this into f(x1), we get f(x1)= f(x0 + h). For our purposes, we are interested in any value of x, so we can replace x0 with just x.  Doing this we can write the slope in terms of x and h is This quotient in terms of f and h we call the difference quotient. Now is a good time to review the previous discussion, perhaps a few times, until you understand it before proceeding. Slope of the Tangent to a Curve The tangent intersects the curve at only one point like so: The notion of a derivative allows us to talk about the slope of a line tangent to a curve at some point x.  This slope is the instantaneous rate of change of the curve at a point (x, f(x)).  (Note, it is common to say curve instead of function, but we are referring to the instantaneous rate of change of the function.) We start with a curve and pass a line through the curve.  This line is called a secant. We will keep one point of intersection fixed and move the other toward the first rotating the line as needed.  Once the two points coincide the line will become tangent to the curve at that point. The curve below is a trace of the function f(x) = x2 + C. The fixed point is x. The varying point begins at x + h. The secant is shown in blue. The slope of the secant is its rise over its run.  The rise is the length of the green dashed line and the run is the length of the red dashed line.  The fixed point is (x, f(x)) and the varying point starts at (x+h, f(x+h)).   We will be moving the second point down the curve closer and closer to the fixed point at (x, f(x)).  As we do this we'll see the slope of the secant get smaller causing it to drop near the fixed point (x, f(x)).  Finally, when the two points coincide, the secant will become a tangent to the curve at the point (x, f(x)). As you move from one graph to the next, h becomes smaller and smaller until it reaches zero.  Notice how the slope of the secant changes along the way.  Once h becomes zero the equation for the slope changes from the ratio of the rise over the run to the limit of this ratio as h approaches zero.   We know this limit to be the derivative of f(x) at x. The slope of the secant changes, causing it to rotate about the fixed point (x, f(x)).   This is shown by the gray dashed lines, and the blue arrow shows the direction it rotates. Notice how the slope of the secant which is now a tangent becomes the limit as h approaches 0 of the difference ratio. The following table is a numeric demonstration of the previous graphs.  Recall that f(x) = x2 + C and we are ignoring C since it plays no part in the difference quotient. NOTE: In the numerator of the difference quotient the constant C is eliminated. f(x+h) = (x+h)2 + C     and    f(x) = x2 + C.    f(x+h) - f(x)   (x+h)2 + C - (x2 + C)   (x+h)2 + C - x2 - C   (x+h)2 - x2 so,   f(x+h) - f(x) = (x+h)2 - x2 I've added more values for h then there are graphs above. The value for x is 4 and is fixed.  I chose 6 as the initial value for h and decremented it by one until it reached 0.   Notice that (x + h) decreases each step until it reaches 4 where the two points coincide. The values for f(x+h) and f(x) are shown so that we can see their difference in the next to last column.  The last column shows the difference ratio which finally becomes 0/0 which is undefined.   This entry would be the slope of the tangent, but this table is using the equation for the slope of the secant not the slope of the tangent; the slope of the tangent must be found by evaluating the limit at this point as h approaches zero. x h x+h f(x+h) f(x) f(x+h)-f(x) (f(x+h)-f(x))/h 4 6 10 100 16 84 14 4 5 9 81 16 65 13 4 4 8 64 16 48 12 4 3 7 49 16 33 11 4 2 6 36 16 20 10 4 1 5 25 16 9 9 4 0 4 16 16 0 #DIV/0! The difference quotient decreases from 14 to 9 and then becomes undefined.  From the graphs we see the slope decreasing as these values indicate.  As h decreases, f(x+h) decreases from 100 to 16 which is the value for f(x) at x = 4.  The numerator decreases from 84 to 0.   The denominator, h, decreases from 6 to 0.   And the difference quotient starts at 14 and decreases to 9.   At this point we have 0/0.  But remember, when we talk about a limit we are NOT interested in what happens at h = 0, but rather, what happens when h is NEAR 0.   This is the beauty of the limit; we can still talk about the limiting value this ratio becomes as h approaches 0 even though at h = 0 the ratio is undefined. The slope of the tangent to f(x) at x = 4 is 8. This number, 8, is the derivative of f(x) at x = 4. In the next section we'll show that we can relax the condition that x is fixed and instead evaluate the limit for the difference quotient of this function.  If we do this we get 2x (shown in an example below). So, in general, the derivative of this function is 2x. This may look strange, but we have evaluated the limit without substituting in a number at the last step. We now have an equation that tells us what the slope of the secant is to that curve for every value of x. So at x = 100, the slope of the tangent to f(x) at x = 100 is 2(100) = 200.  The slope of the tangent at x = -5  is  2(-5) = -10.   One can find the equation of the tangent to f(x) at any x by finding its slope using the derivative of f(x) then using the point slope equation of a line. For example, let C = 0, then at x = 4, y = 16 and the slope is 2(4) = 8 8 = (y - 16)/(x - 4). Solving for y,   y = 8x - 16 We talked about the limit as h approaches zero of the difference quotient. We started to the right of x0 and decreased h so that x1 became smaller until it reached x0.   Next we'll require that the slope of the secant starting from the left of x0 must change to equal the same value for the tangent for the derivative to exist at x0.  In this case h starts as a negative value. The following graph visually demostrates this. In other words the one sided limits must exist and be equal at the point the line is tangent to the curve.  A tangent cannot exist on a corner, nor can it exist on a point of discontinuity, that is, at a hole on the curve. Now is another good time to review the previous discussion, perhaps a few times, until you understand it before proceeding. Definitions and Terminology The derived function of f(x) is this limit: If this limit exists then the function f(x) is differentiable; furthermore, the derivative or finding the derivative of f(x) means to evaluate this limit at x. By saying that a function has a derivative, we are also saying the function is differentiable. The following notation is used to name the derived function.  These notations are spoken: "the derivative with respect to x of f," or the "derivative of f with respect to x." All but the last are commonly used. Situations make one notation more convenient to use over the others.    The last notation, Dx, is found primarily in early papers and texts. When we talk about a derivative at a point x0 in some interval a < x < b, then it is understood that the one sided limits of this difference quotient at x = x0 must exist and be equal for f(x) to be differentiable at x0. Note, If this limit exists then it follows that the f(x) must be continuous.    proof It is not enough for a function to be continuous to have a derivative. The example at the end of this section demonstrates this fact. The derivative is the instantaneous rate of change of f(x) with respect to x at the point x. The word "instantaneous" may seem to imply time. The rate of change is the change in the dependent variable with respect to the change in its independent variable and their dimensions could be any physical quantity, time, distance, position, volume, area, etc. We still use the word instantaneous in all cases with this understanding. Derivatives are used in differential equations to provide mathematical models of physical phenomena. These models provide us great predictive capabilities. Always remember, these "models" are not the phenomena they model, but are mathematical approximations to these phenomena. Examples: Find the derivatives of the following functions. 1) f(x) = x Substitute (x+h) into f(x) to get (x + h). Likewise substitute x into f(x) to get x. x + h - x = h h/h = 1, the limit of a constant is that constant. 2) f(x) = 3x2 Substitute (x+h) into f(x) to get 3(x+h)2 Simplify the numerator. Notice the x2 terms add to 0. Divide the denominator h into the terms of the numerator. The limit of 3h as h tends toward 0 is just 3(0) = 0. Since x does not depend on h, 6x is unaffected by h. The limit is 6x, which is the derivative of 3x2. The one sided limits both evaluate to 6x, so they both exist and are equal. Therefore, this is the derivative of f(x). 3) f(x) = 1/x Substitute into f(x) and combine fractions in the numerator. Use parenthesis to group the numerator separating it from the denominator of the limit quotient. Simplify the numerator.  There is no need to multiply out its denominator. Invert the denominator in the difference quotient and multiply. Cancel the h's and evaluate the limit. Evaluate the limit. The h term becomes zero.  All other terms remain. What remains is the x2 term in the denominator. Recall that f(x) is a hyperbola rotated 900 from the X axis. Also recall this function is asymptotic to both the X and Y axes. It is undefined at x = 0, in fact its one sided limits are opposite in sign and unbounded; there is no fixed number that f(x) can approach as x approaches 0. As x increases positively, y decreases toward zero and as x decreases negatively y increases toward zero.    In either case there is a fixed number that y approaches as x approaches an arbitrary value c. In the domain (∞-, 0) ∪ (0, ∞+) the two sided limits exist about x = c (c ≠ 0) and equals 1/x2. Therefore 1/x2 is the derivative of f(x). Example where f(x) is continuous, but f(x) not differentiable at x = 0 The following graph is f(x) = |x|.     f(x) is defined everywhere. The difference quotient g(x) is defined everywhere except at x=0. The limit of this difference quotient as x increases toward 0 is -1 but as x decreases toward 0 the limit of this difference quotient is 1. Since the one sided limits of g(x) are unequal then the function f(x) can not differentiable at x = 0. From an intuitive perspective f(x) must be "smooth" at x to be differentiable at x. Tangents don't exist at "corners." Remember, this intuitive understanding is not enough mathematically. Yet it can be helpful understanding the concept of a derivative. Let's have another look at the previous graph for f(x) = |x|. We've seen the absolute value function in the algebra topic. |x| = x for positive x and |x| = -x for negative x (and 0 for x = 0). So, the graph for f(x) are the lines shown in red above. {y = x for x > 0 and y = -x for x < 0} From a Calculus perspective, f(x) is continuous. Remember, this statement implies continuity throughout the domain for f(x). At every c, and in particular at c = 0, the limit for f(x) = f(c). There are no breaks/holes in the range of f(x). The difference quotient for this function is g(x) and its graph is shown in green. This difference quotient is the slope for each of two lines for f(x). Notice that g(x) has a hole at x = 0. For x < 0, f(x) = -x, so f(x+h) = -(x+h) and f(x) = -x.    g(x) is therefore (-(x+h) - (-x))/h = -h / h = -1. So as negative h approaches zero the left sided limit of g(x)) is -1. For x > 0, f(x) = x, so f(x+h) = (x + h) and f(x)= x.    g(x) is therefore ((x + h) - x)/h = h/h = 1. So, as positive h approaches zero, the right sided limit of g(x) is 1. The one sided limits of g(x) as h approaches 0 are unequal. So in general f(x) has no derivative. If we remove zero from the domain of f(x), then f(x) is differentiable, and its derivative is -1 for x < 0 and +1 for x > 0. Earlier we claimed that if a function is differentiable then it must be continuous.  The following is a proof of that fact. This proof uses (x - x0) instead of h. To introduce f(x0) it is added then subtracted to f(x).  This is a commonly used techique in proofs to get an expression needed by the proof.  In this case we need the difference quotient. To get the difference quotient we multiply the difference of the first two terms by (x - x0) then divide by (x - x0).  Using the vinculum as a grouping symbol we can immediately write this divisor under the difference of f(x) and f(x0).   Recall that when we multiply and divide by the same quantity it is the same as multiplying by 1. The first limit is the definition of f'(x), the second limit evaluates to 0 which eliminates the first limit. Finally the third limit is that of the constant f(x0) and therefore equals f(x0). To be continuous, the limit of f(x) at x0 must exist and equal f(x0).  So we now find the limit as x approache x0 of all terms of the equation using the product and sum rules for limits. Finally we get the result that the limit of f(x) at x0 equals f(x0) This is another FREE TEMPLATE PRINTABLE presented to you from the Calculus section of K12math.com

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