f(x) f(a) x a Our intuition tells us that the slope of the tangent line to the curve at the point P is m P Q =



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Lecture 6 : Derivatives and Rates of Cange In tis section we return to te problem of finding te equation of a tangent line to a curve, y f(x) If P (a, f(a)) is a point on te curve y f(x) and Q(x, f(x)) is a point on te curve near P, ten te slope of te secant line troug P and Q is given by m P Q Our intuition tells us tat te slope of te tangent line to te curve at te point P is m m P Q m P Q Q P x a x a Wit tis in mind, we make te following definition: Definition Wen f(x) is defined in an open interval containing a, te Tangent Line to te curve y f(x) at te point P (a, f(a)) is te line troug P wit slope provided tat te it exists m x a Example Find te equation of te tangent line to te curve y x at P (1, 1) (Note: Tis is te problem we solved in Lecture by calculating te it of te slopes of te secants At tis point we ave become more efficient wit our calculations of its) m x a f(x) f(a) x a Te equation of te tangent to te curve at P (1, 1) is : Note Te it in te definition above can be rewritten as follows: m x a f(a + ) f(a) Tis it gives te slope of te line tangent to te curve y f(x) at P (a, f(a)) if te it exists Te slope of te tangent line to a curve at a point (wen it exists) is sometimes called te slope of te curve at tat point y-values on te curve near te point are close to corresponding y-values on te tangent line (We will examine tis property more closely wen we get to Linear Approximation) Example Find te equation of te tangent line to te grap of f(x) x + 5x at te point (1, 6) m 0 f(a+) f(a) 1

Derivative of a function at a point a Definition Wen f(x) is defined in an open interval containing a, te derivative of te function f at te number a is f f(a + ) f(a) (a) x a if te it exists Note Te slope of te tangent line to te grap y f(x) at te point (a, f(a)) is te derivative of f at a, f (a) Example Let f(x) x + 5x Find f (a), f () and f ( 1) Equation of te Tangent Line Note tat te equation of te tangent line to te grap of a function f at te point (a, f(a)) is given by (y f(a)) f (a)() Example Find te equation of te tangent line to te grap y x + 5x at te point were x Find te equation of te tangent line to te grap y x + 5x at te point were x 1 Note Wen te derivative of a function f at a, is positive, te function is increasing and wen it is negative, te function is decreasing Wen te absolute value of te derivative is small, te function is canging slowly (a small cange in te value of x leads to a small cange in te value of f(x)) Wen te absolute value of te derivative is large, te function values are canging rapidly (a small cange in x leads to a large cange in f(x))

Some its are easy to calculate wen we recognize tem as derivatives: Example Te following its represent te derivative of a function f at a number a In eac case, wat is f(x) and a? (a) x f(x) a (a) sin x 1 x / x x sin x 1 x / f(x) f(a) x / (1 + ) + (1 + ) f(x) a (1+) +(1+) 0 f(a+) f(a) Velocity If an object moves in a straigt line, te displacement from te origin at time t is given by te position function s f(t), were s is te displacement of te object from te origin at time t Te average velocity of te object over te time interval [t 1, t ] is given by f(t ) f(t 1 ) t t 1 Te velocity (or instantaneous velocity) at time t a is given by te following it of average velocities: v(a) t a f(t) f(a) t a 0 f(a + ) f(a) f (a) Tus te velocity at time t a is te slope of te tangent line to te curve y s f(t) at te point were t a Example Te position function of a stone trown from a bridge is given by s(t) 10t 16t feet (below te bridge) after t seconds (a) Wat is te average velocity of te stone between t 1 1 and t 5 seconds? Wat is te instantaneous velocity of te stone at t 1 second (Note tat speed Velocity ) 3

Different Notation, Rates of cange, x, y If y is a function of x, y f(x), a cange in x from x 1 to x is sometimes denoted by x x x 1 and te corresponding cange in y is denoted by y f(x ) f(x 1 ) Te difference quotient y x f(x ) f(x 1 ) x x 1 is called te average rate of cange of y wit respect to x Tis is te slope of te line segment P Q, were P (x 1, f(x 1 )) and Q(x, f(x )) are on te grap y f(x) Te instantaneous rate of cange of y wit respect to x, wen x x 1, is te it of te slopes of line segments P Q as Q gets closer and closer to P on te grap y f(x); y x 0 x f(x ) f(x 1 ) ( f (x 1 )) x x 1 x x 1 Note tat tis is just te derivative of f(x) wen x x 1 Tus we ave anoter interpretation of te derivative: Te derivative, f (a) is te instantaneous rate of cange of y f(x) wit respect to x wen x a Wen te instantaneous rate of cange is large at x 1, te y-vlaues on te curve are canging rapidly and te tangent as a large slope Wen te instantaneous rate of cange ssmall at x 1, te y-vlaues on te curve are canging slowly and te tangent as a small slope In economics, te instantaneous rate of cange of te cost function (revenue function) is called te Marginal Cost (Marginal Revenue ) Example Te cost (in dollars ) of producing x units of a certain commodity is C(x) 50 + x (a) Find te average rate of cange of C wit respect to x wen te production level is canged from x 100 to x 169 Find te instantaneous rate of cange of C wit respect to x wen x 100 (Marginal cost wen x 100, usually explained as te cost of producing an extra unit wen your production level is 100)

Example Te cost (in dollars ) of producing x units of a certain commodity is C(x) 50 + x (a) Find te average rate of cange of C wit respect to x wen te production level is canged from x 100 to x 169 Solution Te average rate of cange of C is te average cost per unit wen we increase production from x 1 100 tp x 169 units It is given by x y f(x ) f(x 1 ) x x 1 50 + 169 (50 + 100) 169 100 13 10 69 3 69 037 Find te instantaneous rate of cange of C wit respect to x wen x 100 (Marginal cost wen x 100, usually explained as te cost of producing an extra unit wen your production level is 100) Solution Te instantaneous rate of cange of C wen x 100 It is given by x x 100 y f(x) f(100) 50 + x (50 + 100) x 10 x 100 x 100 x 100 x 100 x 100 x 100 ( x 10) x 100 ( x 10)( x + 10) x 100 1 ( x + 10) 1 0 05 5