1. Aim: Basic Differentiation Course: Calculus Do Now: Aim: What are some of the basic rules of differentiation? On what interval(s) is the derivative of f(x) increasing?

2. Aim: Basic Differentiation Course: Calculus Definition of the Derivative The derivative of f at x is provided this limit exists. The derivative f’(x) is a formula for the slope of the tangent line to the graph of f at the point (x,f(x)). The function found by evaluating the limit of the difference quotient is called the derivative of f at x. It is denoted by f ’(x), which is read “f prime of x”.

3. Aim: Basic Differentiation Course: Calculus Alternate Notation for Derivative... read as “the derivative of y with respect to x.”

4. Aim: Basic Differentiation Course: Calculus The derivative of a constant function is 0. That is, if c is a real number, then The Constant Rule y = 7 function derivative f(x) = 0 s(t) = -3 f’(x) = 0 s’(t) = 0

5. Aim: Basic Differentiation Course: Calculus If n is a rational number, then the function f(x) = x n is differentiable and For f to be differentiable at x = 0, n must be a number such that x n-1 is defined on an interval containing 0. The Power Rule f(x) = x 3 function derivative y = 1/x 2 Note:

6. Aim: Basic Differentiation Course: Calculus Model Problems Find the slope of the graph f(x) = x 4 using the power rule when x = 0. f(x) = x 4 f’(x) = 4x 3 f’(0) = 4(0) 3 = 0 Power Rule evaluate d/dx for 0

7. Aim: Basic Differentiation Course: Calculus Model Problems Find the equation of the tangent line to the graph f(x) = x 2 when x = -2. f(x) = x 2 f’(x) = 2x f’(0) = 2(-2) = -4 = m - the slope of tangent line at (2,-4) Power Rule y – y 1 = m(x – x 1 ) Point Slope form y – 4 = -4[x – (-2)] Substitute for y 1,, m, and x 1 y = -4x – 4 Simplify

8. Aim: Basic Differentiation Course: Calculus The Constant Multiple Rule If f is a differentiable function and c is a real number, then cf is also differentiable and y = 2/x functionderivative

9. Aim: Basic Differentiation Course: Calculus The Sum & Difference Rules The derivative of the sum (or difference) of two differentiable functions is differentiable and is the sum (or difference) of their derivatives. Sum Rule Difference Rule f(x) = -x 4 /2 + 3x 3 – 2x functionderivative f’(x) = -2x 3 + 9x 2 – 2 g(x) = x 3 – 4x + 5g’(x) = 3x 2 – 4

10. Aim: Basic Differentiation Course: Calculus Do Now: Aim: What are some of the basic rules of differentiation? Find the equation of the tangent line for f(x) = 4x 3 - 7x 2, at x = 3.

11. Aim: Basic Differentiation Course: Calculus Sine and Cosine The derivatives of the Sine and Cosine Functions: y = 2 sin x function derivative y = x + cos x y’ = 2 cos x

12. Aim: Basic Differentiation Course: Calculus Recall: Finding Slope at Point x, f(x) (x + h,f(x + h)) h f(x + h) – f(x) x, f(x) h f(x + h) – f(x) (x + h,f(x + h)) x, f(x) h f(x + h) – f(x) (x + h,f(x + h)) difference quotient the first derivative of the function

13. Aim: Basic Differentiation Course: Calculus Average Rate of Change (x + h,f(x + h)) h f(x + h) – f(x) x, f(x)

14. Aim: Basic Differentiation Course: Calculus Average Rate of Change The average rate of change of a function f(x) on an interval [a, b] is found by computing the slope of the line joining the end points of the function on that interval. Find avg. rate of change f(x) = x 2 for [-1, 2] (b,(b,f(b)) b - a f(b) – f(a) a, f(a)

15. Aim: Basic Differentiation Course: Calculus Rates of Change Following is graph of f with various line segments connecting points on the graph. The average rate of change of f over the interval [.5, 2] is the slope of which line segment? CD

16. Aim: Basic Differentiation Course: Calculus Rates of Change D = rt If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s = -16t 2 + 100 where s is measured in feet and t is measured in seconds. Find the average velocity over the time interval [1, 1.5]. Average velocity = Position Function g = -32 ft/sec 2 -40 ft/sec

17. Aim: Basic Differentiation Course: Calculus x, f(x) (t + h,s(t + h)) h s(t + h) – s(t) t, s(t) Instantaneous Velocity – Time & Distance t, s(t) h s(t + h) – s(t) (t + h,s(t + h)) t, s(t) h s(t + h) – s(t) (t + h,s(t + h)) Velocity Function the first derivative of the Position Function is g = -32 ft/sec 2 time distance

18. Aim: Basic Differentiation Course: Calculus Model Problems At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by where s is measured in feet and t is measured in seconds. a.When does the diver hit the water? b.What is the diver’s velocity at impact? a. diver hits water when s = 0. Substitute and solve for t. 2 sec.

19. Aim: Basic Differentiation Course: Calculus Model Problems At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by where s is measured in feet and t is measured in seconds. a.When does the diver hit the water? b.What is the diver’s velocity at impact? 2 sec. b. s’(t) = -32t + 16 Power Rule s’(t) = -32(2) + 16 = -48 Substitute 2 from a. -48 feet per second