Lecture 5: Derivatives of Functions PDF

Chapter 4 Derivative of functions In this chapter the concept of the derivative will be developed. This concept is fundamental to all that follows, so study this material carefully. Rule 1: Constant Function If f(x ) = c , where c is any constant then f ( x)...

Chapter 4 Derivative of functions In this chapter the concept of the derivative will be developed. This concept is fundamental to all that follows, so study this material carefully. Rule 1: Constant Function If f(x ) = c , where c is any constant then f ( x)  0 Example 3: Consider the constant function f(x)=5. Solution Applying rule 1, f ( x )  0 If we consider what the function looks like graphically, this result seem reasonable. The function f(x)=5 graphs as horizontal line intersecting the y-axis at (0,5), The slope at all points along such a function equals 0 Rule 2 : Power rule If f(x) = x ,where n is a real number, then f (x )  n x n n 1 Example 4 Consider f(x) = x. This function is the same as f(x) = x1. Applying rule 2 to find the derivative, n=1. Solution f (x )  n x n 1 = 1 x0  5 Example 5 Consider f(x) = x. Applying rule 2 to find the derivative, n=5 Solution f (x )  5 x 51  5 x4 3 Example 6 Consider f(x ) = x2 Rewriting f as f(x) = x 2 3, Applying rule 2 to find the derivative , n = 2 , 3 1 2 2 1 2 x 3 2 Solution f (x )  x 3   3 3 3 3 x Rule 3 : Constant times a function If f(x) = c  g(x), where c is a constant and g is a differentiable function f ( x)  c  g ( x) Example 7 Consider f(x) = 10 x 2. Applying rule 3 to find the derivative 2 , c=10 ,g(x)= x. Solution f ( x)  c  g ( x)  10(2 x ) =20x Example 8 Consider f(x)= 3 x 1  1 Solution This function can be rewritten as f(x) = 3    3x x  3 f (x )  (3)(1)(x 11 )  3x 2  x2 Rule 4: Sum or difference of functions If f(x) = u(x)  v(x), where u and v are differentiable, f ( x)  u( x)  v( x). Example 9 Consider the function f(x) = x  5x. Applying rule 4 to 2 find the derivative of f(x). Solution f can be expressed as f(x)=u(x)-v(x), f ( x)  u( x) - v( x) = 2 x  5 Example 10 Consider f(x)= 5x 4  8x 3  3x 2  x  50. Applying rule 4 to find the derivative of f(x). Solution: f (x )  5(4x 3 )  8(3x 2 )  3(2x )  1  0  20x 3  24x 2  6x  1 Rule 5: Product rule If f(x)=u(x)  v(x),where u and v are differentiable , then f ( x)  u( x)  v( x)  v( x)  u ( x) Example 11 Consider f(x) = (x 2  5)(x  x 3 ). Applying rule 5 to find the derivative of f(x) 2 Solution: ,u(x)= x - 5 and v(x)= x - x3 Therefore , f ( x)  u( x)  v( x)  v( x)  u ( x)  (2 x)( x - x3 )  (1- 3x 2 )( x 2 - 5)  - 5 x 4  18 x 2 - 5 Example 12 In the previous example the function could have been rewritten in the equivalent form f(x)  x 3  x 5  5x  5x 3  x 3  x 5  5x  5x 3  x 5  6x 3  5x Finding the derivative using this form of function does not require the use of rule 5.The derivative is f ( x)  - 5 x 4  18 x 2 - 5 Which the same result is as obtained before. Rule 6: Quotient rule If f(x) = u (x )v (x ) , where u and v are differentiable and v(x)  0, then v ( x )  u ( x )  u ( x ) v (x ) f ( x )  [v ( x )]2 1 Example 13 Consider f(x) = x3 Applying rule 6 to find the derivative of f(x) Solution v (x )  u (x )  u (x ) v (x ) f (x )   x 3 (0)  (1)(3x 2 ) 3x 2  6 [v (x )]2 (x 3 )2 x (3x 2  5) Example 14 Consider f(x) = (1  x 3 ) Applying rule 6 to find the derivative of f(x) with u(x)= (3x  5) and v(x)= (1  x ) 2 3 Solution  (1  x3 )(6 x)  (3x 2  5)(3x 2 ) 6 x  6 x  9 x  15 x 4 4 2 3x 4  15x 2  6x  f ( x)    (1  x3 )2 (1  x3 )2 (1  x 3 )2 Rule 7 : Power of function If f(x) = [u (x )]n ,where u is differentiable function and n is a real number, then f ( x)  n  [u ( x)]n 1  u ( x) Example 15 Consider f(x)  7x 4  5x  9. Applying rule 7 to find the derivative of f(x) Solution: f ( x)  n[u ( x)]  u( x) n 1 1 1 1 ( ) 1 5  (7x  5x  9) 4 2 (28x  5)  (14x  )(7x  5x  9) 2 3 3 4 2 2 Example 16 Consider the function 5  3x  f ( x)   2   1 x  Applying rule 7 to find the derivative of f(x) Solution 3x  (1  x 2 )(3)  (3x)(2 x)  3x 4 3  3x 2  6x 2 4  f ( x)  5  2  =5  2   1 x  (1  x 2 )2  1  x  (1  x ) 2 2 Rule 8 : Exponential functions (base- e) u (x ) If f(x) = e , where u is differentiable ,then f ( x)  u ( x )eu ( x ) Example 17 Consider f(x)= e x. Applying rule 8 to find the derivative of f(x). Solution : f ( x)  u( x)eu ( x ) =1e x =ex x 2 2x Example 18 Consider f(x) =e.Applying rule 8 to find f ( x ). f ( x)  e x 2 x (2 x  2) 2 Solution: Rule 9 : Natural Logarithm Functions If f(x)= ln u(x), where u is differentiable , then u ( x) f ( x)  u ( x) Example 19 Consider the function f(x)=ln x Applying rule 9 to find f ( x ). u( x) Solution : f ( x)  = 1 u ( x) x Rule 10 : Chain rule If y=f(u) is a differentiable function and u=g(x) is a differentiable function ,then dy dy du =  dx du dx Rule 11 : Derivatives of Trigonometric functions f x   f x  f x  f x  d d dx dx sin x cos x sec x sec x tan x cos x  sin x csc x  csc xcot x tan x sec 2 x cot x  csc2 x Note that if u  u (x) f x   f x  f x  f x  d d dx dx sin u cos u.u  sec u sec u  tan u  u  cosu sin u  u cscu  cscu. cot u. u  tan u sec2 u  u cot u  cosec2u  u Example 20 i ) y  sin(x 3  2x )  y   cos(x 3  2x ).(3x 2  2) ii ) y  sec(sin(x 3  2x )) y  sec(sin(x 3  2x )) tan(sin(x 3  2x )) cos(x 3  2x )(3x 2  2) Higher - order derivatives Example 22 Given f ( x)  x 6  2 x 5  x 4  3x 3  x 2  x  1 Find all additional higher –order derivatives Solution f ( x)  x 6  2 x 5  x 4  3x 3  x 2  x  1 f ( x)  30 x 4  40 x3  12 x 2  18 x  2 f ( x)  120 x3  120 x 2  24 x  18 f (4) (x)  360 x 2  240 x  24 f (5) ( x)  720 x  240 f (6) ( x)  720 f (7) ( x)  0 All additional higher –order derivatives will also equal 0.