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**Introduction to Derivatives**

y Tangent line at (c, f(c)) Secant line between (a, f(a)) and (c, f(c)) (a, f(a)) Definitions (informal): 1. A secant line is a line connecting two points of a curve. 2. A tangent line at a point of a curve is a line that only touches that point of the curve, locally speaking. (c, f(c)) x y = f(x) We can always calculate the slope between two points (recall ). Therefore, we can always calculate the slope of a secant line because it connects two points of the curve. For example the slope of the secant line shown above is m = On the other hand, we can’t find the slope of the tangent line since even though it also touches two points of the curve, only (c, f(c)) is given, the other one is not! Let alone those tangent lines that really touch exactly one point of the curve (see diagram on the right)! Can we find the slope of the tangent line at a point given that we don’t know any other points? The answer is YES. This is when we need to use the idea called limiting process—to find the slope of a tangent line at a given point (c, f(c)) of a function, we can find the slope between the fixed point (c, f(c)) and some movable point, say, (x, f(x)). As (x, f(x)) moving closer and closer to (c, f(c)), we calculate the slopes of the secant lines between (c, f(c)) and (x, f(x)), and see what the limit is, i.e., what the values of the slopes of the secant lines are getting closer to? msec (x, f(x)) (c, f(c)) mtan

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y –6 –4 –2 x O 2 4 6 8 10 –8 12 14 16 Let (1, 1) be fixed and let the other points vary, what are slopes of the following secant lines: Shown: Between (4, 16) and (1, 1): Between (3, 9) and (1, 1): Between (2, 4) and (1, 1): Between (0, 0) and (1, 1): Not Shown: Between (1.1, 1.21) and (1, 1): Between (1.01, ) and (1, 1): Conclusion: The slope of the tangent line must be: What is the equation of the tangent line at (1, 1)? (Hint: use y – y1 = m(x – x1)) y = x2 (1, 1) y –6 –4 –2 x O 2 4 6 8 10 –8 12 14 16 y = x2 (2, 4) Now, let (2, 4) be fixed and let the other points vary, what are slopes of the following secant lines: Shown: Between (4, 16) and (2, 4): Between (3, 9) and (2, 4): Between (1, 1) and (2, 4): Between (0, 0) and (2, 4): Not Shown: Between (2.1, 4.41) and (2, 4): Between (2.01, ) and (2, 4): Conclusion: The slope of the tangent line at (2, 4) must be: What is the equation of the tangent line at (2, 4)?

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**provided that this limit exists. **

Definition (Formal): The tangent line to the graph of a function y = f(x) at a point P = (c, f(c)) is defined as the line containing the point P whose slope is provided that this limit exists. If mtan exists, an equation of the tangent line at (c, f(c)), by the point-slope form (i.e., y – y1 = m(x – x1)), is y – f(c) = mtan(x – c) Therefore, without calculating the slopes of the secant lines of the points near (2, 4), we can just use: where, in this case, f(x) = x2, c = 2, and f(c)) = 4. msec (x, f(x)) (c, f(c)) mtan y –6 –4 –2 x O 2 4 6 8 10 –8 12 14 16 y = x2 (2, 4)

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**Provided that this limit exists.**

y –4 –8 2 4 6 8 10 y = x3 (1, 1) –6 –2 x O Let (1, 1) be fixed and let the other points vary, what are slopes of the following secant lines: Shown: Between (2, 8) and (1, 1): Between (0, 0) and (1, 1): Not Shown: Between (1.1, 1.331) and (1, 1): Between (1.01, ) and (1, 1): Conclusion: The slope of the tangent line must be: What is the equation of the tangent line at (1, 1)? (Hint: use y – y1 = m(x – x1)) Using the definition: Q: What does this have to do with derivative? A: What we did is the derivative of the function at the point! Definition: The Derivative of a Function f at a Number c: Let y = f(x) denote a function f, and if a real number c is in the domain of f, the derivative of f at c. denoted by f(c), read as “f prime of c,” is defined as Provided that this limit exists.

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**As we can see, the formula is really the same as the formula . So,**

Furthermore, if an equation of the tangent line at (c, f(c)) can be written as: y – f(c) = mtan(x – c) it can be written as: How to find the derivative of a function f at c, i.e., f(c)? Example: If f(x) = x2 + 2x – 4, find f(3). Step 1: Find f(c) if it’s not given. Step 2: Use Step 3: Find the limit: Whatever you obtain in step 3 is f(c). Note: when you try to find the , it will be a [0/0] indeterminate form, in almost every case. Examples: Find the derivative at the indicated x-value or at the indicated point. 1. f(x) = x2 + 2 at x = –1 2. f(x) = x2 – 3 at (2, 1) mtan =

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**Step 0: Find f(c) if it’s not given. **

How to find an equation of the tangent line of a function f(x) at a given x-value c or a given point (c, f(c))? Step 0: Find f(c) if it’s not given. Step 1: Find the slope of the tangent line using the formula Step 2: Use point-slope form of a line y – y1 = m(x – x1) by substituting c into x1, f(c) into y1 and f (c) into m. Examples: Find an equation of the tangent line of the given function at the x-value or at the given point. 1. f(x) = 2x2 + x – 2 at x = 1 2. f(x) = x3 – 2 at (2, 6) Determining whether the derivative of f at a number should be positive, negative or zero when a graph is given: When we are given a graph, we know that to determine whether a function value f(c) is positive, negative, or zero, all we need to do is to ask whether the point (c, f(c)) is above, below or on the x-axis, respectively. For example, f(–3) is positive because the point (–3, f(–3)) is above the x-axis; f(4) is negative because the point (4, f(4)) is below the x-axis, f(7) is zero because the point (7, f(7)) is on the x-axis. –1 –2 –3 1 2 3 4 5 6 7 However, to determine the derivative of f at a number c, i.e., f (c), is positive, negative, or zero, we need to ask ourselves this question: If we can construct a tangent line at (c, f(c)), what is the slope of this tangent line? For example, if we construct a tangent line at (–3, f(–3)), what is the slope—positive, negative, or zero? Obviously it’s positive! Therefore, f (–3) is positive. Determine whether the following derivatives are positive (+), negative (–) or zero (0): f (–3) = (+) f (–1) = ___ f (1) = ___ f (3) = ___ f (4) = ___ f (7) = ___

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EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute.

EXAMPLE 1 Find a positive slope Let (x 1, y 1 ) = (–4, 2) = (x 2, y 2 ) = (2, 6). m = y 2 – y 1 x 2 – x 1 6 – 2 2 – (–4) = = 4 6 2 3 = Simplify. Substitute.

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