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5.1 DERIVATIVES OF EXPONENTIAL FUNCTIONS

5.1 DERIVATIVES OF EXPONENTIAL FUNCTIONS. By: Rafal, Paola , Mujahid. RECALL:. y=e x (exponential function) LOGARITHM FUNCTION IS THE INVERSE EX1: y=log 4 x y=4 x Therefore y=e x y=log e x. IN THIS SECTION:.

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5.1 DERIVATIVES OF EXPONENTIAL FUNCTIONS

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  1. 5.1 DERIVATIVES OF EXPONENTIAL FUNCTIONS By: Rafal, Paola , Mujahid

  2. RECALL: • y=ex (exponential function) • LOGARITHM FUNCTION IS THE INVERSE • EX1: y=log4 x y=4x • Therefore y=ex y=log e x

  3. IN THIS SECTION: • The values of the derivative f’(x) are the same as those of the original function y=ex • THE FUNCTION IS ITS OWN DERIVATIVE f(x)=ex f’(x)=ex • “e” is a constant called Euler’s number or the natural number, where e= 2.718

  4. The product, quotient, and chain rules can apply to exponential functions when solving for the derivative. f(x)=e g(x) Derivative of composite function: f(x)=e g(x) f’(x)= e g(x) g’(x) by using the chain rule

  5. Key Points • f(x)= ex , f’(x)= ex • Therefore, y= ex has a derivative equal to itself and is the only function that has this property. • The inverse function of y=ln x is the exponential function defined by y= ex.

  6. Example 1: Find the derivative of the following functions. a) y= e 3x+2 b) y= ex2+4x-1 y’= g’(x) (f(x)) y’= g’(x) (f(x)) y’=(3)(e 3x+2 ) y’= (2x+4)(ex2+4x-1 ) y’= 3 e 3x+2 y’= 2(x+2)(ex2+4x-1 )

  7. You can also use the product rule and quotient rule when appropriate to solve. • Recall: Product rule: f’(x)= p’(x)(q(x)) + p(x)(q’(x)) Quotient rule: f’(x)=p’(x)(q(x)) – p(x)(q’(x)) ___________________________ q(x)2

  8. Example 2: Find the derivative and simplify • f(x)= X2e2x f’(x)= 2(x)(e2x ) + (X2 ) (2)(e2x ) use product rule f’(x)= 2x e2x + 2 X2 e2xsimplify terms f’(x)=2x(1+x) e2x factor out 2x

  9. f(x)= ex ÷ x f’(x)= (1) (ex )(x) – (ex )(1) ÷ X2use quotient rule f’(x)= x ex - ex ÷ X2simplify terms f’(x)= (x-1) (ex ) factor out ex ______ X2

  10. Example 3: Determine the equation of the line tangent to f(x) = ex÷x2 , where x= 2. Solution: Use the derivative to determine the slope of the required tangent. f(x) = ex÷x2 f(x)=x-2exRewrite as a product f’(x)= (-2x-3)ex+ x-2 (1)exProduct rule f’(x)= -2ex ÷ x3 + ex÷ x2Determine common denominator

  11. f’(x)=-2ex ÷ x3 + xex÷ x3Simplify f’(x)= -2ex + xex÷ x3Factor f’(x)= (-2+ x) ex÷ x3 When x=2, f(2) = e2÷ 4. When x=2, f’(2)=0 . so the tangent is horizontal because f’(2)=0. Therefore, the equation of the tangent is f(2) = e2÷ 4

  12. Example 4: The number, N, of bacteria in a culture at time t, in hours, is N(t)=1000(30+e-). • What is the initial number of bacteria in the culture? • Determine the rate of change in the number of bacteria at time t. • What is happening to the number of bacteria in the culture as time passes?

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