Presentation on theme: "2.6 The Derivative By Dr. Julia Arnold using Tan’s 5th edition Applied Calculus for the managerial, life, and social sciences text."— Presentation transcript:
2.6 The Derivative By Dr. Julia Arnold using Tan’s 5th edition Applied Calculus for the managerial, life, and social sciences text
What is the derivative of something? The derivative of a function f(x) is, mathematically speaking, the slope of the line tangent to f(x) at any point x. It is also called “the instantaneous rate of change” of a function. It can be equated with many real world applications such as; Velocity which is speed such as miles per hour In business, the derivative is called marginal, such as the marginal Cost function, etc.
We call a line which intersects a graph in two points a secant line. It is easy to find the slope of that line. But it takes limits in order to find the slope of a tangent line which touches the graph at only one point. The green line is a secant line because it crosses the blue graph more than once. In particular we focus on the two points illustrated. For animation you need to be connected to the internet.
A tangent to a graph.
To find the slope of the tangent line to a graph f(x) we use the following formula: First let’s see how this formula equates with the slope of a tangent line. Notation: The derivative of f(x) is denoted by the following forms: or
(x,f(x)) (x+h,f(x+h)) x x+h The slope of the secant line would be By decreasing h a little each time we get closer and closer to the slope of the tangent line. By using limits and letting h approach 0 we get the actual slope of the tangent line.
The difference quotient measures the average rate of change of y with respect to x over the interval [x,x+h] In a problem pay attention to that word average. If it is there then you do not use limits.
Differentiability and Continuity If a function is differentiable at x = a, then it is continuous at x = a. However, continuous functions may contain points at which the function is not differentiable. These are points that create sharp points graphically, or vertical tangent lines. In this graph there is an abrupt change at (a,f(a)) The slope at (a,f(a)) is undefined.
Example 1:Let Find the derivative f’ of f. Find the point on the graph of f where the tangent line to the curve is horizontal. Find an equation of the tangent line at the point (2, 16).
Substitution of numerator. Multiplying out Combining like terms Factor out h Cancel Let h = 0. Done. Find the derivative f’ of f.
Find the point on the graph of f where the tangent line to the curve is horizontal. The slope of the tangent line anywhere on the curve is given by the formula we just found. f’(x)=2x+6 We know that the slope of a horizontal line is 0, thus Let f’(x) = 0 and find the x that causes that. 0=2x + 6 -6 = 2x -3=x So the x value where the tangent line would be horizontal is -3. For the y value substitute into the original formula f(x) =(-3) 2 +6(-3)=9-18=-9. The point is (-3,-9).
Find an equation of the tangent line at the point (2, 16). The slope of the tangent line at this point would be found by using the formula we found in the first part “2x+6” which would make the slope to be 2(2)+6=10 Using the slope 10 and the point (2,16) and the point slope formula we get Y – 16= 10(x-2) Y – 16 = 10x – 20 Y =10x-4
Find the average rate of change for f(x) = x 2 - 3x + 4 from x = 2 to x = 3, from x = 2 to x = 2.5, from x = 2 to x = 2.1. When finished find the instantaneous rate of change at x = 2. Solution: Use the derivative formula without doing the limit. Thus simplify just the part of the formula.
So f(x) = x 2 - 3x + 4 average rate of change formula simplifies to 2x - 3 + h. First Interval x = 2 to x = 3 h is the difference between the two values or 3-2=1 Second Interval x = 2 to x = 2.5 h =.5 Third Interval x = 2 to x = 2.1 h =.1 Since the second number in the interval is actually x + h in the formula, that makes x = 2 in this problem. First Interval x = 2 and h = 1. Substitute into 2x - 3 + h and the average rate of change is 2(2) - 3 + 1 = 2 Second Interval x = 2 and h =.5 Substitute 2(2)-3 +.5 = 1.5 Third Interval x = 2 and h =.1 Substitute 2(2)-3 +.1 = 1.1
As you can see as we get closer to x = 2 the slopes are decreasing toward 1. For the instantaneous rate of change, we use the limit as h approaches 0 to obtain 2x - 3 at x = 2 which is 1. Similar problems to the ones illustrated here will be on Test 1.