Prep Book Chapter 5 – Definition pf the Derivative Looking to calculate the slope of a tangent line to a curve at a particular point. Use formulas to construct functions that we then calculate a limit for. The limit we calculate IS the value of the SLOPE OF THE TANGENT LINE. Two formulas based on y/x

= Definition 1 for slope of tan. line. Slope Definition 1 y x f(x) - f(a) m = = x - a f f(x) - f(a) x - a lim x-->a Q(x, f(x)) [f(x) - f(a)] = Definition 1 for slope of tan. line. P(a, f(a)) (x - a) x a

ex: Find the equation of the tangent line to the curve y = x2 @ point P = (1,1) f(a) a = 1, f(a) = 1 f(x) - f(a) x - a lim x-->a x2 - 1 (x + 1)(x - 1) (x - 1) = lim x-->1 = lim x-->1 x - 1 = lim x-->1 (x + 1) = 2 = m Point-Slope Form: [y - y1 = m(x - x1)] = [y - 1 = 2(x - 1)] y - 1 = 2x - 2 y = 2x - 1 = equation of the tangent line

= Definition 2 for slope of tan. line. Slope Definition 2 m = y x = f(a+h) - f(a) h f f(a+h) - f(a) h lim h-->0 Q(a+h), f(a+h)) [f(a+h) - f(a)] = Definition 2 for slope of tan. line. P(a, f(a)) h (a + h) a

ex: Find the equation of the tangent line to the hyperbola y = 3 ex: Find the equation of the tangent line to the hyperbola y = @ point P = (3, 1) x f(a) = 1 a = 3 a+h = 3+h 3 (3+h) = lim h-->0 3 - (3+h) (3+h) h - 1 f(a+h) - f(a) h lim h-->0 = lim h-->0 h = lim h-->0 -h (3+h) h = lim h-->0 -h (3+h) x h 1 = lim h-->0 -1 (3+h) = -1 3 1

2.6 cont’d - Velocities Ex: Suppose a ball is dropped from a position 450m high. The equation of its motion is (s) = f(t) = 4.9t2 What is the velocity of the ball after 5 seconds? How fast is the ball traveling when it hits the ground? s (5, 122.5) 122.5 t 5

Calc 2.8 - The Derivative as a Function f(a+h) - f(a) h f’(a) = lim h-->0 f(x+h) - f(x) h f’(x) = lim h-->0 so… If we continuously vary x, f’(x) will also vary…. so if we look at the original function f, then f’(x) is the slope of function f at a particular point x… so if we match up a particular point x on function f with the value of the slope of f at that point x, we have point (x, f’(x))… which means we can PLOT A NEW CURVE using x and using f’(x) as y-values…. which means we can now refer to f’(x) as a separate function…(that can be graphed just like f(x) can)…. 10

ex: f(x) f’(x) 11

ex: f(x) = x3 - x f’(x) = 3x2 - 1 1 ex: f(x) = x f’(x) = 2 x 12

2.8 continued… Differentiation - The process of calculating a derivative. For standard notation y = f(x), derivative notation may be: f ’(x) y ’ dy/dx (Leibniz notation) d/dx f(x) Definition of “Differentiable” - A function is differentiable at a # ‘a’ if f ’(a) exists. It is differentiable on an open interval if it is differentiable at every number in the interval… (a,b) , [a,b], (a,), etc…

Continuity vs. Differentiability If f is differentiable at ‘a’, then f is continuous at ‘a’. If f is continuous at ‘a’, f is NOT necessarily differentiable at ‘a’. Non-Differentiable Characteristics: Kinks Discontinuities Vertical Tangents

f (x) is a function and may have its own derivative (f )  = f  The Second Derivative f (x) is a function and may have its own derivative (f )  = f  Leibniz notation: d dx dy = d2y dx2 ex: For f(x) = x3 - x and f (x) = 3x2 - 1, find and interpret f (x)… f(x+h) - f(x) h f (x) = lim h-->0

3(x + h)2 - 1 - (3x2 - 1) h lim = f (x) = 3x2 - 1 3(x2 + 2xh + h2) - 1 - (3x2 - 1) h lim h-->0 = 3x2 + 6xh + 3h2 - 1 - 3x2 + 1 h lim h-->0 = 6xh + 3h2 h lim h-->0 = lim 6x + 3h h-->0 = 6x = f (x) Interpretation: Any derivative is a rate of change - therefore the “derivative of a derivative” is “a rate of change of a rate of change”… the second derivative is the rate at which the first derivative is changing.

f(x) f(x) f(x) f(x) f(x) f(x) 17

Derivative Chain of Position Velocity Acceleration s(t) s’(t) = v(t) s’’(t) = v’(t) = a(t) ds dt dv dt d2s dt2 or Leibniz Notation: Acceleration = instantaneous rate of change of velocity.