LU8 Derivative I_student.ppt - LU8 Derivative I The Derivative Four Basic Rules of Differentiation The General Power Rule The Product and Quotient Rules

LU8 Derivative I_student.ppt - LU8 Derivative I The...

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Unformatted text preview: LU8 Derivative I The Derivative Four Basic Rules of Differentiation The General Power Rule The Product and Quotient Rules: Higher-Order Derivatives The Chain Rule Differentiation of Exponential and Logarithmic Functions 8.1 Four Basic Rules of Differentiation Rule 1: Derivative of a Constant We will use the notation d f ( x ) dx to mean “the derivative of f with respect to x at x.” Rule 1: Derivative of a constant d c 0 dx The derivative of a constant function is equal to zero. Rule 1: Derivative of a Constant We can see geometrically why the derivative of a constant must be zero. The graph of a constant function is a straight line parallel to the x axis. Such a line has a slope that is constant with a value of zero. Thus, the derivative of a constant must be zero as well. y f(x) = c x Rule 2: The Power Rule Rule 2: The Power Rule If n is any real number, then d n n 1 x nx dx Rule 2: The Power Rule Practice Examples 1: If f(x) = x, then If f(x) = x8, then If f(x) = x5/2, then d f ( x ) x 1 x1 1 x 0 1 dx f ( x ) d 8 x 8 x 8 1 8 x 7 dx f ( x ) d 5/2 5 5/2 1 5 3/2 x x x dx 2 2 Rule 2: The Power Rule Practice Examples 2: Find the derivative of f ( x ) x Rule 2: The Power Rule Practice Examples 3: 1 Find the derivative of f ( x ) 3 x Rule 3: Derivative of a Constant Multiple Function Rule 3: Derivative of a Constant Multiple Function If c is any constant real number, then d d cf ( x ) c f ( x ) dx dx Rule 3: Derivative of a Constant Multiple Function Practice Examples 4: Find the derivative of f ( x ) 5 x 3 d f ( x ) 5 x 3 dx d 3 5 x dx 5 3x 2 15 x 2 Rule 3: Derivative of a Constant Multiple Function Practice Examples 5: 3 Find the derivative of f ( x ) x Rule 4: The Sum Rule Rule 4: The Sum Rule d d d f ( x ) g ( x ) f ( x ) g ( x ) dx dx dx Rule 4: The Sum Rule Practice Examples 6: Find the derivative of f ( x ) 4 x 5 3x 4 8 x 2 x 3 d f ( x ) 4 x 5 3x 4 8 x 2 x 3 dx d 5 d 4 d 2 d d 4 x 3 x 8 x x 3 dx dx dx dx dx 4 5 x 4 3 4 x 3 8 2 x 1 0 20 x 4 12 x 3 16 x 1 Rule 4: The Sum Rule Practice Examples 7: Find the derivative of t2 5 g (t ) 3 5 t Applied Example 8: Conservation of a Species A group of marine biologists at the Neptune Institute of Oceanography recommended that a series of conservation measures be carried out over the next decade to save a certain species of whale from extinction. After implementing the conservation measure, the population of this species is expected to be N (t ) 3t 3 2t 2 10t 600 (0 t 10) where N(t) denotes the population at the end of year t. Find the rate of growth of the whale population when t = 2 and t = 6. How large will the whale population be 8 years after implementing the conservation measures? Applied Example 8: Conservation of a Species Solution The rate of growth of the whale population at any time t is given by N (t ) 9t 2 4t 10 In particular, for t = 2, we have 2 N (2) 9 2 4 2 10 34 And for t = 6, we have 2 N (6) 9 6 4 6 10 338 Thus, the whale population’s rate of growth will be 34 whales per year after 2 years and 338 per year after 6 years. Applied Example 8: Conservation of a Species Solution The whale population at the end of the eighth year will be 3 2 N 8 3 8 2 8 10 8 600 2184 whales 8.2 The General Power Rule If the function f is differentiable and h(x) = [f(x)]n then (n, a real number), d n n 1 h( x ) f ( x ) n f ( x ) f ( x ) dx The General Power Rule Practice Examples 9: Find the derivative of G ( x ) x 2 1 Solution 1/2 Rewrite as a power function: G ( x ) x 2 1 Apply the general power rule: 1/2 d 1 2 2 G( x ) x 1 x 1 2 dx 1/2 1 2 x 1 2 x 2 x x2 1 The General Power Rule Practice Examples 11: Find the derivative of f ( x ) 1 4x 2 7 2 8.3 The Product and Quotient Rules Rule 5: The Product Rule The derivative of the product of two differentiable functions is given by d f ( x ) g ( x ) f ( x ) g ( x ) g ( x) f ( x ) dx Rule 5: The Product Rule Practice Examples 13: Find the derivative of f ( x ) 2 x 2 1 x 3 3 f ( x ) 2 x 2 1 d 3 d 3 x 3 x 3 2 x 2 1 dx dx 2 x 2 1 3 x 2 x 3 3 4 x 6 x 4 3x 2 4 x 4 12 x x 10 x 3 3x 12 Rule 5: The Product Rule Practice Examples 14: Find the derivative of f ( x ) x 3 x 1 Rule 5: The Product Rule Practice Examples 10: 5 Find the derivative of f ( x ) x 2 2 x 3 Rule 6: The Quotient Rule The derivative of the quotient of two differentiable functions is given by d f ( x) g ( x) f ( x) f ( x) g ( x) 2 dx g ( x) g ( x ) g x 0 Rule 6: The Quotient Rule Practice Examples 15: Find the derivative of x f ( x) 2x 4 d d 2 x 4 ( x) x 2 x 4 dx dx f ( x ) 2 2 x 4 2 x 4 1 x 2 2 2 x 4 2x 4 2x 2 x 4 2 4 2 x 4 2 Rule 6: The Quotient Rule Practice Examples 16: Find the derivative of x2 1 f ( x) 2 x 1 Rule 6: The Quotient Rule Practice Examples 12: Find the derivative of 2x 1 f ( x ) 3 x 2 3 Applied Example 17: Rate of Change of DVD Sales The sales ( in millions of dollars) of DVDs of a hit movie t years from the date of release is given by 5t S (t ) 2 t 1 Find the rate at which the sales are changing at time t. How fast are the sales changing at: ✦ The time the DVDs are released (t = 0)? ✦ And two years from the date of release (t = 2)? Applied Example 17: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing at time t is given by d 5t S (t ) 2 dt t 1 t 2 1 5 5t 2t t 2 1 2 5t 5 10t t 2 1 2 2 2 51 t2 t 2 1 2 Applied Example 17: Rate of Change of DVD Sales Solution The rate of change at which the sales are changing when the DVDs are released (t = 0) is 2 5 1 0 5 1 S (0) 5 2 2 2 1 0 1 That is, sales are increasing by $5 million per year. Applied Example 17: Rate of Change of DVD Sales Solution The rate of change two years after the DVDs are released (t = 2) is 2 5 1 2 5 1 4 15 3 S (2) 0.6 2 2 2 25 5 4 1 2 1 That is, sales are decreasing by $600,000 per year. 8.4 Higher-Order Derivatives The derivative f ′ of a function f is also a function. As such, f ′ may also be differentiated. Thus, the function f ′ has a derivative f ″ at a point x in the domain of f if the limit of the quotient f ( x h ) f ( x ) h exists as h approaches zero. The function f ″ obtained in this manner is called the second derivative of the function f, just as the derivative f ′ of f is often called the first derivative of f. By the same token, you may consider the third, fourth, fifth, etc. derivatives of a function f. Higher-Order Derivatives Practice Examples 18: Find the third derivative of the function f(x) = x2/3 and determine its domain. Solution 2 2 1 2 We have f ( x ) x 1/3 and f ( x ) x 4/3 x 4/3 3 3 3 9 So the required derivative is 2 4 7/3 8 7/3 8 x x 9 3 27 27 x 7/3 The domain of the third derivative is the set of all real numbers except x = 0. f ( x ) Higher-Order Derivatives Practice Examples 19: Find the second derivative of the function f(x) = (2x2 +3)3/2 Applied Example 20: Acceleration of a Maglev The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function s = 4t 2 (0 t 10) What is the maglev’s acceleration after 30 seconds? Applied Example 20: Acceleration of a Maglev The distance s (in feet) covered by a maglev moving along a straight track t seconds after starting from rest is given by the function s = 4t 2 (0 t 10) What is the maglev’s acceleration after 30 seconds? Solution The velocity of the maglev t seconds from rest is given by ds d v 4t 2 8t dt dt The acceleration of the maglev t seconds from rest is given by the rate of change of the velocity of t, given by d d ds d 2 s d a v 2 8t 8 dt dt dt dt dt or 8 feet per second per second (ft/sec2). 8.5 The Chain Rule Deriving Composite Functions 2 Consider the function h ( x ) x x 1 2 To compute h′(x), we can first expand h(x) 2 2 h( x ) x x 1 x 2 x 1 x 2 x 1 x 4 2 x 3 3x 2 2 x 1 and then derive the resulting polynomial h( x ) 4 x 3 6 x 2 6 x 2 But how should we derive a function like H(x)? 2 H ( x ) x x 1 100 Deriving Composite Functions 2 Note that H ( x ) x x 1 100 is a composite function: H(x) is composed of two simpler functions f ( x) x 2 x 1 g ( x ) x100 and So that H ( x ) g f ( x ) f ( x ) 100 2 x x 1 100 We can use this to find the derivative of H(x). Deriving Composite Functions To find the derivative of the composite function H(x): We let u = f(x) = x2 + x + 1 and y = g(u) = u100. Then we find the derivatives of each of these functions du f ( x ) 2 x 1 dx and dy g (u ) 100u 99 du The ratios of these derivatives suggest that dy dy du 100u 99 2 x 1 dx du dx Substituting x2 + x + 1 for u we get 99 dy 2 H ( x ) 100 x x 1 2 x 1 dx Rule 7: The Chain Rule If h(x) = g[f(x)], then h( x) d g f ( x) g f ( x) f ( x) dx Equivalently, if we write y = h(x) = g(u), where u = f(x), then dy dy du dx du dx The Chain Rule for Power Functions Many composite functions have the special form h(x) = g[f(x)] where g is defined by the rule g(x) = xn so that h(x) = [f(x)]n (n, a real number) In other words, the function h is given by the power of a function f. Examples: 2 h( x ) x x 1 100 H ( x) 1 3 3 5 x G( x) 2 x 2 3 Applied Problem 21: Arteriosclerosis Arteriosclerosis begins during childhood when plaque forms in the arterial walls, blocking the flow of blood through the arteries and leading to heart attacks, stroke and gangrene. Suppose the idealized cross section of the aorta is circular with radius a cm and by year t the thickness of the plaque is h = g(t) cm then the area of the opening is given by A = (a – h)2 cm2 Further suppose the radius of an individual’s artery is 1 cm (a = 1) and the thickness of the plaque in year t is given by h = g(t) = 1 – 0.01(10,000 – t2)1/2 cm Applied Problem 21: Arteriosclerosis Then we can use these functions for h and A h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = (1 – h)2 to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule: dA dA dh f (h ) g (t ) dt dh dt 1 2 1/2 2 (1 h )( 1) 0.01 10,000 t ( 2t ) 2 0.01 t 2 (1 h ) 1/2 10,000 t 2 0.02 (1 h )t 10,000 t 2 Applied Problem 21: Arteriosclerosis Then we can use these functions for h and A h = g(t) = 1 – 0.01(10,000 – t2)1/2 A = f(h) = (1 – h)2 to find a function that gives us the rate at which A is changing with respect to time by applying the chain rule: dA dA dh f (h ) g (t ) dt dh dt 1 2 1/2 2 (1 h )( 1) 0.01 10,000 t ( 2t ) 2 0.01 t 2 (1 h ) 1/2 10,000 t 2 0.02 (1 h )t 10,000 t 2 Applied Problem 21: Arteriosclerosis For example, at age 50 (t = 50), h g (50) 1 0.01(10,000 2500)1/2 0.134 So that dA 0.02 (1 0.134)50 0.03 dt 10,000 2500 That is, the area of the arterial opening is decreasing at the rate of 0.03 cm2 per year for a typical 50 year old. 8.5 Differentiation of the Exponential and Logarithmic Functions Rule 8 Derivative of the Exponential Function The derivative of the exponential function with base e is equal to the function itself: d x e e x dx Examples 22 Find the derivative of the function f ( x ) x 2e x Examples 22 Find the derivative of the function f ( x ) x 2e x Solution Using the product rule gives d 2 x 2 d x x d x e x e e x2 dx dx dx x 2e x e x (2 x ) f ( x ) xe x ( x 2) Examples 23 Find the derivative of the function g (t ) et 2 3/2 Rule 9 Chain Rule for Exponential Functions If f(x) is a differentiable function, then d f ( x) f ( x) e e f ( x ) dx Examples 24 Find the derivative of the function f ( x ) e 2 x 3x 3x Find the derivative of the function y e 2 t 2 t Find the derivative of the function g (t ) e 2x Find the derivative of the function y xe 2 x et Find the derivative of the function g (t ) t e e t Examples 24 Find the derivative of the function f ( x ) e 2 x Solution d f ( x ) e 2x dx e 2 x (2) 2x 2e 2 x Rule 10 Derivative of the Natural Logarithm The derivative of ln x is d 1 ln x dx x ( x 0) Examples 25 Find the derivative of the function f ( x ) x ln x Solution d d f ( x ) x (ln x ) ln x ( x ) dx dx 1 x ln x (1) x 1 ln x Examples 26 ln x x Find the derivative of the function g ( x ) ln x Solution Rule 11 Chain Rule for Logarithmic Functions If f(x) is a differentiable function, then d f ( x ) ln f ( x ) dx f ( x) [ f ( x ) 0] Examples 27 Find the derivative of the function f ( x ) ln( x 2 1) Solution d 2 x 1 f ( x ) dx 2 x 1 2x 2 x 1 Examples 28 Find the derivative of the function y ln[( x 2 1)( x 3 2)6 ] Solution y ln[( x 2 1)( x 3 2)6 ] ln( x 2 1) ln( x 3 2)6 ln( x 2 1) 6ln( x 3 2) d 2 d 3 ( x 1) ( x 2) dy dx dx 6 dx x2 1 x3 2 2x 3x 2 2 6 3 x 1 x 2 2x 18 x 2 2 3 x 1 x 2 ...
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