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Lecture

# Derivatives

12 Pages
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School
University of Ottawa
Department
Mathematics
Course
MAT1320
Professor
Kirill Zaynullin
Semester
Fall

Description
Calculus and Vectors – How to get an A+ 2.1 Derivative Function A Derivative Function B Differentiability Given a function y = f (x), the derivative A function y = f (x)is called differentiable atxif f '(xexists. function of f is a new function called f (f A function y = f (x)is differentiable over an open interval(a,b)if prime), defined at x by: the function is differentiable at every number in that interval. Note: The domain of derivative function f is a subset of the f x ) = limf x+ )h ( f x h→0 h domain of the original function f : D f ' D f. So a function is defined over D fbut is differentiable overD f ' C Interpretations of Derivative Function D Notations and Reading 1. The slope of the tangent line to the graph y'= f '(x) [Lagrange Notation] of y = f (x)at the pointP(a, f (a)is given by Read: “y prime” or “f prime at x” m = f '(a). dy d 2. The instantaneous rate of change in the = f (x) [Leibnitz Notation] variable y with respect to the variable x , dx dx Read: “dee y by dee x” where y = f (x, at x= a is given by: IRC = f '(a. dy f a ) = dx x=a Read: “f prime at a, dee y by dee x at x equals a” E First Principles Ex 1. Use first principles to differentiate each function. Differentiation is the process to find the a) f (x) = 2x − x derivative function for a given function. f (x + h) = 2(x + h)−(x + h) First Principles is the process of f x+ h) − f x) 2(x +h ) (x + h)3− [2x− x3] differentiation by computing the limit: f x )= lim = lim h→0 h h→ 0 h 3 3 f x+ )h ( f x = lim 2(x+ h) 2 x − lim (x+ h) − x f x ) = lim h→0 h h→0 h h→0 h h[( + h)2+ ( + h)x + x2] 2 = lim2 − lim = − x h→0 h→0 h ∴ f '(x) = 2−3x2 − 3 c) f (x) = ax + b b) f x = 2 x −3 f (x h) = 2 f'x )= lim f x + h)− f x) = lim a x + h)+ b − ax+ b (x + ) h→0 h h→0 h −3 − 3 − a x + h) +b − ax+ b a x + h)+ b + ax + b f x +h )− f x) (x + h)2 x2 = lim f x )= lim = lim h→0 h a x + h)+ b + ax + b h→0 h h→0 h 1 1 a x + h)+ b − (ax+ b) 2 − 2 2 2 = h→0 (x+ h) x x − (x+ h) h( a x+ h) +b + ax+ b ) = − 3h→0 = −3lh→0 2 2 h h x+ h) x = lim ah = lim a 2 2 2 h→0 h( a x+ h) +b + ax+ b ) h→0 a x+ h) +b + ax +b = − 3lim x − [x +2 xh+ h ] = −3lim − 2x − h h→0 h x + h)2x2 h→ 0(x+ h)2 x2 a a = = − x 6 a x +0) + b+ ax +b 2 ax + b = − 3 2 2 = 3 x x x ∴ f 'x = a 6 ∴ f'( )= 3 2 ax + b x 2.1 Derivative Function ©2010 Iulia & Teodoru Gugoiu - Page 1 of 2 Calculus and Vectors – How to get an A+ G Non-Differentiability Notes: A function is not differentiable at x = a if 1. If a function f is not continuous at x = athen the function f is f '(adoes not exist. not differentiable at x = a. 2. If a function f is continuous at x = a then the function f may be or not differentiable at x = a. H Differentiability Point If the function y = f (x)is differentiable at x = a then the tangent line at P(a, f (a))is unique and not vertical (the slope of the tangent line is not ∞ or −∞ ). I Corner Point Ex 2. Find the numbers x where the function y = f (x)(see the P(a, f (a))is a corner point if there are two graph below) is not differentiable and explain why. distinct tangent lines at P , one for the left- hand branch and one for the right-hand branch. For example: ⎧f ( ), x < a f( ) = ⎨ 1 and f1'(a) ≠ f2'(a) ⎩f2(x), x > a J Infinite Slope Point P(a, f (a)is a infinite slope point if the tangent line at P is vertical and the function is increasing or decreasing in the neighborhood at the of the point P . f '(a) = ∞or f '(a) = −∞ The function y = f (x) is not differentiable at: x = −6 (infinite break) x = −2 (P(−2,4) is a corner point) x = 0 (jump discontinuity) x = 2 (P(2,4) is a cusp point) K Cusp Point x = 4 (removable discontinuity) P(a, f (a))is a cusp point if the tangent line x = 6 (P(6,4) is an infinite slope point) at P is vertical and the function is increasing on one side of the point P and decreasing on the other side. f '(a) = DNE Reading: Nelson Textbook, Pages 65-72 Homework: Nelson Textbook: Page 73 #1, 6, 7b, 9, 14, 16, 19 2.1 Derivative Function ©2010 Iulia & Teodoru Gugoiu - Page 2 of 2 Calculus and Vectors – How to get an A+ 2.2 Derivative of Polynomial Functions A Power Rule Ex 1. For each case, differentiate. Consider the power function: y = x , x,n∈R . a) (1)'= (x )'= 0x0−1= 0 Then: b) (x)'= (x )'=1(x1−1) =1(x ) =1 c) (x )'= 5(x5−1) = 5x4 (x )'= nx n−1 ' d n n−1 d) ⎛1 ⎞ =(x −1)' ( 1) −1 1= −x −2= − 1 x = nx ⎜ x⎟ 2 dx ⎝ ⎠ x ' Some useful specific cases: e) ⎜ 1 ⎟ = (x−7)'= (−7)x −7 1= −7 x−8 = − 7 ⎝ x7 ⎠ x 8 (1)' 0 1 1−1 −1 f)( x)' =x )' = 1 x 2 = 1 x 2 = 1 1 = 1 ( )' 1 2 2 2 1 2 x x 2 ( x )' 1 1 1 1−1 1 −2 1 1 2 x g) ( x)'= (x )'= x3 = x 3 = = 3 3 2 3 2 3 x3 3 x 3 3 −1 4 3 3 −1 3 3 3 h) ( x )'= (x )'= x 4 = x 4 = = 4 4 1 4 4x 4x 4 π π −1 i)(x )'=πx B Constant Function Rule Ex 2. Find each derivative function: Let consider a constant function: f (x) = c, c∈R . a) (−2)'= 0 Then: b) ( π )'= 0 (c)'= 0 d dx c = 0 C Constant Multiple Rule Ex 3. Differentiate each expression: Let consider g(x) = cf (x. Then: a) − 2x 3 (−2x )'= (−2)(x )'= (−2)(3)x 3−1 = −6x 2 [cf( )]'= cf'( ) −3 d d b) 2 [cf( )] = c f( ) x dx dx ' (cf)'= cf' ⎛ −3 ⎞ = ( 3 −2)' ( 3)( x−2 )' ( 3)( 2) − −1= 6 −3 = 6 ⎜ 2 ⎟ 3 ⎝ x ⎠ x c) 2x 1 1 ( 2x)'= 2( x)'= 2 = 2 x 2x D Sum and Difference Rules Ex 4. Differentiate. a) f (x) = 2 − 3x + 4x − 3x7 [ ( ) ± g ( )]f '( ) g '( ) f '(x) = (2 −3x + 4x −3x )' d d d [ ( ) ± g( )]= f( ) ± g(x) = (2)'+(−3x)'+(4x )'+(−3x )' dx dx dx ( f ± g)'= f ' g' = 0 + (−3)(x)'+4(x )'−3(x )' = −3+ 4(2)x −3(7)x 6 = −3+8x − 21x 6 ∴ f '(x) = −3+8x − 21x 6 2.2 Derivative of Polynomial Functions ©2010 Iulia & Teodoru Gugoiu - Page 1 of 4 Calculus and Vectors – How to get an A+ 1 2 4 x − 3 x b) g(x) = − 2 + 5 c) h( ) = x x x x ' ' ' ⎛1 2 4 ⎞ −1 −2 −5 ⎛ x − 3 x⎞ ⎛ x 3x ⎞ g'( ) ⎜ − 2 + 5 ⎟ = ( − 2 + 4 )' h'( )= ⎜ ⎟ =⎜ − ⎟ ⎝x x x ⎠ ⎝ x ⎠ ⎝ x x ⎠ − −2 −5 = ( )' ( 2x )' (4 )' 1−1 1−1 −1 −2 −1 −2 −2 −3 −6 = (x2 −x 3 )= (x 2 − x 3 )= (x 2 )− ( 3 )' = − 1( ) ( 2)( 2)x + 4( 5) 1 4 20 ⎛− 1⎞ −−1 ⎛ − 2⎞ −2−1 ⎛− 1⎞ −3 ⎛ −2 ⎞ −5 = − + − = ⎜ ⎟x 2 −⎜ ⎟x 3 = ⎜ ⎟x 2 − ⎜ ⎟x 3 x 2 x3 x 6 ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ∴ g x( )= − 1 + 4 − 20 −1 2 1 2 x2 x 3 x6 = 3 + 3 5 = − + 3 2 2 x 3 x 2 x x 3x x 1 2 ∴ h'( )= − + 2x x 3 2 3 x x E Tangent Line Ex 5. Find the equation of the tangent line at the point To find the equation of the tangent line at the point 1 P(1,0)to the graph of y = f (x) = x − . P(a.f (a)): x 1. Find derivative function f '(x). 1 f x ) = x + 2. Find the slope of the tangent line using: x2 m = f '(a) 2 m = f (1) = 2(1)+1/1 = 3 3. Use the slope-point formula to get the equation of the tangent line: y −0 = 3(x −1) ∴ y = 3x −3 y − f (a) = m(x −a) Ex 6. Find the equation of the tangent line of the 3 2 Ex 7. Find the points on the graph of y = f (x) = 2x −3x +1 slope m = 2 to the graph of y = f (x) = 2 x. Graph where the tangent line is horizontal. the function and the tangent line. 1 1 f '(x) = 6x −6x = 6x(x −1) f'( ) 2 = 2 x x f '(x) = m 1 1 1 m = 0 (horizontal tangent) f'( )= 2 ⇒ = 2 ⇒ x= ⇒ y =2 =1 6x x − 1)= 0 x 4 4 P(1/4,1) ⇒ y− 1 2( x −1/4) x = 0 ⇒ y =1 ⇒ A(0,1) ∴ =y 2 x+ 1/2 x =1 ⇒ y= 0 ⇒ B (1,0) The tangent line is horizontal at A(0,1)and B(1,0). y = 2*x+1/2x) y 3 y 2 2 1 1 x −2 −1 1 2 x −3 −2 −1 1 2 3 −1 −1 −2 −2 −3 2.2 Derivative of Polynomial Functions ©2010 Iulia & Teodoru Gugoiu - Page 2 of 4 Calculus and Vectors – How to get an A+ F Normal Line Ex 8. Find the equation of the normal line to the curve 2 If m T is the slope of the tangent line, then slope of y = f (x) = x + at P(1,3) . the normal line m is given by: x N 1 2 m = − f '( )= − 2 N m x T m T = f(1) = −1 m = − 1 =1 N m T y −3 =1( −1) ∴ =y x + 2 Ex 9. Analyze the differentiability of each function. 2/3 b) y = f (x) = x 3 The function f is continuous over R . a) y = f (x) = x 2 1 2 −1 2 − 2 The function f is continuous over R . f '(x) = x3 = x 3 = 1 1 2 3 3 3 x 1 −1 1 − 1 f'( ) = (x 3)'= x 3 = x 3 = f '(x)does not exist at x = 0 . 3 3 3 2 3 x The function f is not differentiable at x = 0 . f '(x)does not exist at x = 0. Therefore the − As x → 0 , f '(x) → −∞ . function f is not differentiable at x = 0 . As x → 0 +, f '(x) → ∞ . As x → 0 , f '(x) → ∞ . The point O(0,0) is a infinite The point O(0,0) is a cusp point. slope point. 3 y y 3 2 2 1 x 1 −3 −2 −1 1 2 3 x −1 −3 −2 −1 1 2 3 −1 −2 −3 −2 −3 y c) y = f (x) =| x −3| 4 ⎧ x −3, x ≥ 3 ⎧1, x > 3 3 f x) = ⎨ ⇒ f x( ) = ⎨ 2 3− ,x x < 3 −1, x < 3 ⎩ ⎩ 1 lim f '(x) = −1 ≠ lim f '(x) =1 x x→3− x→3 + −2 −1 1 2 3 4 5 6 7 −1 f 'does not exist at x = 3 . Therefore, f is not −2 differentiable at x = 3 . The point P(3,0) is a corner −3 point (see the figure to the right side). −4 2.2 Derivative of Polynomial Functions ©2010 Iulia & Teodoru Gugoiu - Page 3 of 4 Calculus and Vectors – How to get an A+ H Differentiability for piece-wise defined Ex 10. Analyze the differentiability of each function at x = 1 . functions ⎪ x2 , x ≤1 Let consider the piece-wise defined function: a) f( ) = ⎨ ⎪ 2 , x >1 ⎧ 1 (x), x < a ⎩ ⎪ f (x) = ⎨, x = a ⎪ lim f (x) =1, lim f (x) = 2, f (1) =1 ⎩ f2(x), x > a x→1− x→1 + The function f is not continuous at x =1 . Therefore, the The function f is differentiable at x = a if: function f is not differentiable at x =1 . See the figure (a) the function is continuous at x = a below. (b) f '(a) = f '(a) (the slope of the tangent line for 1 2 y the left branch is equal to the slope of the 4 tangent line for the right branch). 3 2 1 x −3 −2 −1 1 2 3 −1 ⎪ x2 , x ≤1 ⎪ x2 , x ≤1 b) a) f x) = c) a) f x) = ⎪ x, x >1 ⎪ 2x−1, x >1 ⎩ ⎩ lim f (x) =1, lim f (x) =1, f (1) =1 lim f (x) =1, lim f (x) =1, f (1) =1 x→1− x→1 + x→1− x→1 + The function is continuous at x =1 . The function is continuous at x =1 . ⎧ 2 , x < 1 ⎧2 , x < 1 f'( ) = ⎨
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