Differentiation Rules an Formulas Professor D. Olles December 1, 01 1 Te Definition of te Derivative Consier a function y = f(x) tat is continuous on te interval a, b]. Ten, te slope of te secant line passing troug (a, f(a)) an (b, f(b)) is given by m s = f(b) f(a) b a If we let te secon point be variable, meaning tat it is allowe to move along te curve for all x, ten te slope of te secant line can be written as a function. m a (x) = f(x) f(a) x a As te point (x, f(x)) approaces te point (a, f(a)) we see tat x a an te secant line becomes a tangent line at te instant wen te two points meet. Tus, f(x) f(a) m t m s (x) x a x a x a Since x is approacing some constant value a, m t is not a function of x rater, it is also constant. Now consier te same senario but were te two points are bot variable. Tat is, we ave (x, f(x)) an (x +, f(x + )). Te slope of te secant line can now be written in te form m s (x) = f(x + ) f(x) In tis form, we are consiering tat te point (x +, f(x + )) will approac (x, f(x)). Tis means tat x + x an terefore 0. Te slope of te tangent line is now f(x + ) f(x) m t (x) Notice tat ere, te slope of te tangent line is written as a function of x. Tis is ue to te fact tat bot initial an terminal points were variable an te istance between tem is wat was allowe to be cange. 1
Since te slope of te tangent line represents te instantaneous rate of cange of our function f for every x, we may write te erivative of f as: Te Rules f f(x + ) f(x) (x) = m t (x).1 Te Sum an Difference Rules Let f an g be continuous functions. Ten, x f(x) ± g(x)] = f (x) ± g (x) Proof of te sum rule: Let m(x) = f(x) + g(x) be a continuous function. Ten, m (x) m(x + ) m(x) f(x + ) + g(x + )] f(x) + g(x)] f(x + ) + g(x + ) f(x) g(x) f(x + ) f(x) + g(x + ) g(x) ] f(x + ) f(x) g(x + ) g(x) + f(x + ) f(x) g(x + ) g(x) + lim = f (x) + g (x) Proof of te ifference rule: Let m(x) = f(x) g(x) be a continuous function. Ten, m (x) m(x + ) m(x) f(x + ) g(x + )] f(x) g(x)] f(x + ) g(x + ) f(x) + g(x) f(x + ) f(x) g(x + ) + g(x)
] f(x + ) f(x) g(x + ) + g(x) + ] f(x + ) f(x) g(x + ) g(x) f(x + ) f(x) g(x + ) g(x) lim = f (x) g (x). Te Scalar Multiple Rule x cf(x)] = cf (x) Let m(x) = cf(x) be a continuous function were f is ifferentiable an c is a constant. Ten, m (x) m(x + ) m(x) cf(x + ) cf(x) cf(x + ) f(x)] = c lim f(x + ) f(x) = cf (x).3 Te Prouct Rule x f(x)g(x)] = f (x)g(x) + f(x)g (x) Let m(x) = f(x) g(x) be a continuous function were f an g are ifferentiable. Ten, m m(x + ) m(x) (x) f(x + ) g(x + ) f(x) g(x) f(x + ) g(x + ) f(x + ) g(x) + f(x + ) g(x) f(x) g(x) f(x + )g(x + ) g(x)] + f(x + ) f(x)]g(x) 3
f(x + )g(x + ) g(x)] + ] f(x + ) f(x)]g(x) f(x + )g(x + ) g(x)] f(x + ) f(x)]g(x) + lim g(x + ) g(x) f(x + ) lim = f(x)g (x) + f (x)g(x) f(x + ) f(x) + lim lim g(x).4 Te Quotient Rule ] f(x) = f (x)g(x) f(x)g (x) x g(x) g(x)] Let m(x) = f(x) g(x) be a continuous function were f an g are ifferentiable an g(x) 0. Ten, m (x) m(x + ) m(x) f(x+) g(x+) f(x) g(x) f(x+)g(x) f(x)g(x+) g(x+)g(x) f(x + )g(x) f(x)g(x + ) g(x + )g(x) f(x + )g(x) f(x)g(x) + f(x)g(x) f(x)g(x + ) g(x + )g(x) f(x + ) f(x)]g(x) + f(x)g(x) g(x + )] g(x + )g(x) f(x + ) f(x)]g(x) f(x)g(x + ) g(x)] g(x + )g(x) ] f(x + ) f(x)]g(x) f(x)g(x + ) g(x)] g(x + )g(x) g(x + )g(x) f(x + ) f(x) lim g(x) g(x + )g(x) lim = f (x) g(x) g(x)] f(x) g(x)] g (x) = f (x)g(x) g(x)] f(x)g (x) g(x)] = f (x)g(x) f(x)g (x) g(x)] f(x) g(x + )g(x) lim g(x + ) g(x) 4
.5 Te Cain Rule x f(g(x))] = f (g(x))g (x) Let m(x) = 3 Derivative Formulas 3.1 Te Constant Function x c] = 0 Let f(x) = c were c is some constant. Ten, f(x + ) = c. f (x) f(x + ) f(x) c c 0 0 = 0 3. Power Functions x xn ] = nx n 1 Let f(x) = x n were n is some constant. f (x) f(x + ) f(x) (x + ) n x n x n + nx n 1 + n(n 1) x n 1 + + nx n 1 + n x n nx n 1 + n(n 1) x n 1 + + nx n 1 + n ( ) nx n 1 + n(n 1) x n 1 + + nx n + n 1 5
( nx n 1 + ) n(n 1) x n 1 + + nx n + n 1 = nx n 1 + 0 + 0 + + 0 + 0 = nx n 1 3.3 Te Absolute Value Function x x = x x ; x 0 Let f(x) = x. f (x) f(x + ) f(x) x + x ( x + x ) ( x + + x x + + x (x + ) (x) ( x + + x ) x + x + x ( x + + x ) x + ( x + + x ) (x + ) ( x + + x ) x + x + + x = x + 0 x + 0 + x = x x = x x ) 6
3.4 Exponential Functions Let f(x) = a x. 3.5 Logaritmic Functions x ax ] = a x ln a f (x) f(x + ) f(x) a x+ a x a x a a x a x (a 1) = a x lim a 1 a 1 Since lim = ln a we ave f (x) = a x ln a x log a x] = 1 x ln a Verifie: Let y = log a x an sow tat y x = 1 x ln a. y = log a x a y = a log a x a y = x x ay ] = x x] Using te erivative of te exponential function an implicit ifferentiation, we ave: a y ln a y x = 1 y x = 1 a y ln a Since a y = x from above, we ave: y x = 1 x ln a 7
3.6 Trigonometric Functions x sin x] = cos x Let f(x) = sin x. f (x) f(x + ) f(x) sin (x + ) sin x By te sum rule of sines, sin (x + ) = sin x cos + sin cos x x cos x] = sin x Let f(x) = cos x. f sin cos x + sin x cos sin x (x) ] sin cos x sin x cos sin x + sin cos x sin x cos sin x + lim sin cos x sin x (cos 1) + lim sin cos 1 = cos x lim + sin x lim sin cos 1 Since lim = 1 an lim = 0 we ave: f (x) = cos x1] + sin x0] = cos x f (x) f(x + ) f(x) cos (x + ) cos x By te sum rules of cosines, cos (x + ) = cos x cos sin x sin f cos x cos sin x sin cos x (x) cos x cos cos x sin x sin cos x cos cos x 8 ] sin x sin
x tan x] = sec x Verification: Let f(x) = tan x = sin x cos x. x cot x] = csc x cos x cos cos x sin x sin lim cos x(cos 1) sin x sin lim cos 1 sin = cos x lim sin x lim = cos x0] sin x1] = sin x f (x) = tan x] = x x ] sin x cos x Using te quotient rule, we ave: f (x) = Verification: Let f(x) = cot x = cos x sin x. cos x cos x] sin x sin x] cos x] = cos x + sin x cos x = 1 cos x = sec x f (x) = cos x ] cot x] = x x sin x Using te quotient rule, we ave: f (x) = sin x sin x] cos x cos x] sin x] = sin x cos x sin x = sin x + cos x sin x = 1 sin x = csc x 9
x sec x] = sec x tan x Verification: Let f(x) = sec x = 1 cos x. x csc x] = csc x cot x Verification: Let f(x) = csc x = 1 sin x. x f (x) = f (x) = cos x0] 1 sin x] cos x] = sin x cos x = 1 sin x cos x cos x = sec x tan x sin x0] 1 cos x] sin x] = cos x sin x = 1 cos x sin x sin x = csc x cot x 3.7 Inverse Trigonometric Functions sin 1 x ] = 1 1 x Verification: Let y = sin 1 x an sow tat y x = 1 1 x. Since sin y = x = x 1 = OP P HY P sin y = x sin y] = x x x] cos y y x = 1 y x = 1 cos y we can set up a rigt triangle. 10
cos y = 1 x 1 y x = 1 1 x x cos 1 x ] = 1 1 x x x x x tan 1 x ] = 1 1+x cot 1 x ] = 1 1+x sec 1 x ] = 1 x x 1 csc 1 x ] = 1 x 1 x 3.8 Hyperbolic Trigonometric Functions 11