Differentiation of Eponential Fnctions The net derivative rles that o will learn involve eponential fnctions. An eponential fnction is a fnction in the form of a constant raised to a variable power. The variable power can be something as simple as or a more comple fnction sch as + 5. Basic Eponential Fnction = b, where b > 0 and not eqal to 1 Eponential Fnction with a fnction as an eponent = b g ( The derivative of an eponential fnction wold be determined b the se of the chain rle, which was covered in the previos section. In reviewing the derivative rles for eponential fnctions we will begin b looking at the derivative of a fnction with the constant raised to a simple variable. Derivative of an eponential fnction in the form of = b If = b where b > 0 and not eqal to 1 then the derivative is eqal to the original eponential fnction mltiplied b the natral log of the base. ( ln b = b Eample 1: = 5. Soltion: Since o have a constant raised to the variable, the derivative wold be eqal to the original fnction mltiplied b the natral log of the base, which is 5. = 5 = ln 5 5 This derivative rle can be simplified when the base of the eponential fnction is eqal to e. The derivative involves the natral log of the base. However, if the base is eqal to e then the Gerald Manahan SLAC, San Antonio College, 008 1
natral log of the base can be redced to the vale of 1. (See or logarithms formla sheet for a fll list of logarithm properties. This wold simplif the derivative to the original fnction itself. = e = = ( ln ( 1 = e ee e Derivative of an eponential fnction in the form of = e If = e then the derivative is simpl eqal to the original fnction of e. Eample : = e. Soltion: Since the base of the eponential fnction is eqal to e the derivative wold be eqal to the original fnction. = e = e Now lets sa o are given the fnction = b g and are asked to find its derivative. In this case, o will need to se the chain rle to determine the derivative. To see how the chain rle wold be sed we will rewrite this fnction as the composition of the fnctions f( and g(. We will begin b letting eqal the eponent of g( = g = b = b g Now we will let f( eqal Gerald Manahan SLAC, San Antonio College, 008
f = = b f = b Therefore, g( = f g = b. Now recalling the chain rle from the last section we can determine the rle for finding the derivative of an eponential fnction. Chain rle: d d If = f g( then = f g ( g ( In the chain rle formla, f g( = f ( derivative of f( and g(. f = b ( ln f = b b Now sbstitting g( back for gives s: ( ln f = b b ( ln f g = b b for this problem. So what we need to do is find the g Sbstitting this derivative into the chain rle formla will then give o the derivative rle for finding the derivative of eponential fnctions. g ( = f g = b d d ( ln = f g g = g bb g Gerald Manahan SLAC, San Antonio College, 008
As with the previos derivative rles for eponential fnctions, this rle can be simplified to d g( = e g ( if the base is eqal to e. d Derivatives of g ( b and g ( e If d = = d g ( = b then ( ln g bb g If g = e then d = = d g e g Eample : = 5 Soltion: Here o have a constant raised to a fnction so o will se the derivative rle g( = ln bb g = 5 ( ln = g b b g ( ln 5 5 D ( = 1 ( ln 5 5 ( = ( ln 5 5 ( 6 = = 6 ln5 5 Gerald Manahan SLAC, San Antonio College, 008 4
Eample 4: = 6e Soltion: Since the base of the eponential fnction in this problem is e o can se the g( derivative rle = e g = 6e g = e g = e D 6 = 6e = 18e Eample 5: = 4 e Soltion: This problem involves the prodct of two fnctions, one a power fnction and the other an eponential fnction. To find the derivative o will have to appl a combination of the prodct rle, the power rle, and the eponential rle. Step 1: Appl the prodct rle. = 4 e ( 4 ( 4 = D e + e D Step : Appl the power and eponential rles. ( 4 ( 4 = D e + e D 1 ( 4 e D ( ( e ( 4 = + 1 ( 4 e ( ( e ( 1 = + ( 4 ( e ( 4 ( e ( 1 = + Gerald Manahan SLAC, San Antonio College, 008 5
Eample 5 (Contined: Step : Simplif the derivative ( 4 ( 4 ( 1 = e + e 4 ( 16 ( e ( e ( 1 = + ( = 4e 4 + Gerald Manahan SLAC, San Antonio College, 008 6