Exact rate of change

20 - 1 Exact Rate of Change!

Try the quiz at the bottom of the page! go to quiz In the introduction section, we talked about the average rate of change of a function. What happens as you let the "h" value approach zero? Why, of course you are talking about a limit! Look what happens to the line as h approaches zero!

The red curve is the function and the blue line is the slope at a given point! Does this line cross the function twice like the previous section? No way dude! It only crosses the function once. What is a line that crosses a circle exactly once called? If you don't know, ask Mrs. Parker!! That's right, it is a tangent line!! What you have found by letting h approach zero, is the slope of the tangent line!! Important topic approaching!!

Lookie! lookie!!

The slope of the tangent line =

Which is denoted f ' (x) and is called the derivative of f!!

Sample problem Find f'(x) if f(x) = x².

Therefore, f'(x) = 2x for any point on the curve of x². Now find the slope at the point x = 2 and then at x = -3 Piece of cake. simply find the following: f ' (2) = 2(2) = 4 f ' (-3) = 2(-3) = -6 This means that the slope at x = 2 is 4 and the slope at x = -3 is -6. Both of these make sense when you consider at x = 2 the graph of the parabola is increasing and at x = -3 it is decreasing!

To find f '(x) by the definition becomes very tedious work. Therefore, we have some simple formulas to find the derivatives of various functions. Each of these formulas will be proved in class.

1) If f(x) = c, where c is a constant, then f '(x) = 0.
This makes sense when you consider a constant graph is a horizontal line and all horizontal lines have a slope of zero!

2) If f(x) = xⁿ, then f '(x) = nx^{n - 1}.

a) f(x) = x², f '(x) = 2x

b) f(x) = x⁵, f '(x) = 5x⁴.

c) f(x) = x^-4, f '(x) = -4x^-5.

d) f(x) = x^1/2, f '(x) = 1/2(x)^-1/2.

Easy to do. Bring the power out front and decrease the power by one!

3) If f(x) = cxⁿ, then f '(x) = cnx^{n - 1}.

a) f(x) = 3x², f '(x) = 6x.

b) f(x) = -4x³, f '(x) = -12x².

4) If f(x) = p(x) + q(x), then f '(x) = p '(x) + q '(x).

a) f(x) = 3x² + 4x + 1, f '(x) = 6x + 4 (remember, the derivative of a constant is zero!)

b) f(x) = x³ + 4x² - 5x + 3, f '(x) = 3x² + 8x - 5.

Find the slope at x =1

f '(1) = 3(1)² + 8(1) - 5 = 3 + 8 - 5 = 6

Problems to try

Find the first derivative of each and find the slope at x = 2

1) f(x) = 2x⁴

2) f(x) = 5x³ - 4x² + 3x -7

3) f(x) = 1/x²

4) f(x) =

5) f(x) =

Answers at the bottom of the page!

Here are the answers!

1) f '(x) = 8x³, f '(2) = 8(2)³ = 8(8) = 64

2) f '(x) = 15x² - 8x + 3, f '(2) = 15(2)² - 8(2) + 3 = 15(4) - 16 + 3 = 47

3) f(x) = x^-2, f '(x) = -2x^-3, f '(2) = -2(2)^-3 = -2/8

5) f '(x) = 4x² + 3/x² - 8/x³ , f '(2) = 4(2)² + 3/(2)² - 8/(2)³ = 15 3/4

Current quizaroo # 20a

a) 2x + 4; b) 2x + 4 + h; c) 2x + 3; d) 2x + 3 + h; e) 0
: a) 8; b) 12; c) 4; d) 10; e) 6
3) Find the derivative of f(x) = 9x^2/3.: a) 6x^5/3; b) 6x; c) 6; d) 6x^-1/3; e) 0

4) Find the derivative of f(x) = 4/x³.: a) 12x²; b) 12/x²; c) -12x²; d) -12/x²; e) -12/x⁴

5) Find the slope of the function f(x) = 3x² + 2x - 1 at the point (1, 4)

a) 26

b) 8

c) 4

d) 55

e) 10

click here for answers!!