By: Rafal, Paola, Mujahid

 y=e x (exponential function)  LOGARITHM FUNCTION IS THE INVERSE  EX1: y=log 4 xy=4 x  Therefore y=e x y=log e x

 The values of the derivative f’(x) are the same as those of the original function y=e x  THE FUNCTION IS ITS OWN DERIVATIVE f(x)=e x f’(x)=e x  “e” is a constant called Euler’s number or the natural number, where e= 2.718

 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

 f(x)= e x, f’(x)= e x  Therefore, y= e x 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= e x.

 Example 1: Find the derivative of the following functions. a) y= e 3x+2 b) y= e x2+4x-1 y’= g’(x) (f(x)) y’= g’(x) (f(x)) y’=(3)(e 3x+2 ) y’= (2x+4)(e x2+4x-1 ) y’= 3 e 3x+2 y’= 2(x+2)(e x2+4x-1 )

 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

Example 2: Find the derivative and simplify a) f(x)= X 2 e 2x f’(x)= 2(x)(e 2x ) + (X 2 ) (2)(e 2x ) use product rule f’(x)= 2x e 2x + 2 X 2 e 2x simplify terms f’(x)=2x(1+x) e 2x factor out 2x

b) f(x)= e x ÷ x f’(x)= (1) (e x )(x) – (e x )(1) ÷ X 2 use quotient rule f’(x)= x e x - e x ÷ X 2 simplify terms f’(x)= (x-1) (e x ) factor out e x ______ X 2

Example 3: Determine the equation of the line tangent to f(x) = e x ÷x 2, where x= 2. Solution: Use the derivative to determine the slope of the required tangent. f(x) = e x ÷x 2 f(x)=x -2 e x Rewrite as a product f’(x)= (-2x -3 )e x + x -2 (1)e x Product rule f’(x)= -2e x ÷ x 3 + e x ÷ x 2 Determine common denominator

f’(x)=-2e x ÷ x 3 + xe x ÷ x 3 Simplify f’(x)= -2e x + xe x ÷ x 3 Factor f’(x)= (-2 + x) e x ÷ x 3 When x=2, f(2) = e 2 ÷ 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) = e 2 ÷ 4