3 DIFFERENTIATION RULES
DIFFERENTIATION RULES 3.6 Derivatives of Logarithmic Functions In this section, we: use implicit differentiation to find the derivatives of the logarithmic functions and, in particular, the natural logarithmic function.
DERIVATIVES OF LOGARITHMIC FUNCTIONS An eample of a logarithmic function is: y = log a An eample of a natural logarithmic function is: y = ln
DERIVATIVES OF LOG FUNCTIONS It can be proved that logarithmic functions are differentiable. This is certainly plausible from their graphs.
DERIVATIVES OF LOG FUNCTIONS Formula 1 Proof d 1 (log ) a d ln a Let y = log a. Then, a y =. Differentiating this equation implicitly with respect to, using Formula 5 in Section 3.4, we get: So, dy 1 1 d y a ln a ln a a y dy (ln a) 1 d
DERIVATIVES OF LOG FUNCTIONS Formula 2 If we put a = e in Formula 1, then the factor on the right side becomes ln e = 1 and we get the formula for the derivative of the natural logarithmic function log e = ln. d d (ln ) 1
DERIVATIVES OF LOG FUNCTIONS By comparing Formulas 1 and 2, we see one of the main reasons why natural logarithms (logarithms with base e) are used in calculus: The differentiation formula is simplest when a = e because ln e = 1.
DERIVATIVES OF LOG FUNCTIONS Eample 1 Differentiate y = ln( 3 + 1). To use the Chain Rule, we let u = 3 + 1. Then y = ln u. So, dy dy du 2 1 du 1 2 3 (3 ) 3 3 d du d u d 1 1
DERIVATIVES OF LOG FUNCTIONS Formula 3 In general, if we combine Formula 2 with the Chain Rule, as in Eample 1, we get: d 1 du d g '( ) (ln u) or ln g( ) d u d d g( )
DERIVATIVES OF LOG FUNCTIONS Eample 2 Find d ln(sin ). d Using Formula 3, we have: d d 1 d ln(sin ) (sin ) sin d 1 cos cot sin
DERIVATIVES OF LOG FUNCTIONS Eample 3 Differentiate f ( ) ln. This time, the logarithm is the inner function. So, the Chain Rule gives: f '( ) 1 12 d 2 (ln ) (ln ) d 1 1 1 2 ln 2 ln
DERIVATIVES OF LOG FUNCTIONS Eample 4 Differentiate f() = log 10 (2 + sin ). Using Formula 1 with a = 10, we have: f '( ) d log 10 (2 d sin ) 1 d (2 sin )ln10 d cos (2 sin )ln10 (2 sin )
DERIVATIVES OF LOG FUNCTIONS E. g. 5 Solution 1 Find d 1 ln. d 2 d d ln 1 2 1 1 2 d d 1 2 2 2 1 ( 1) 1 2 ( 1 2 2) 1 2 1 2 ( 2 1) 5 ( 1)( 2) 2( 1)( 2)
DERIVATIVES OF LOG FUNCTIONS E. g. 5 Solution 2 If we first simplify the given function using the laws of logarithms, then the differentiation becomes easier: d 1 d 1 ln ln( 1) 2 ln( 2) d 2 d This answer can be left as written. 1 1 1 1 2 2 However, if we used a common denominator, it would give the same answer as in Solution 1.
DERIVATIVES OF LOG FUNCTIONS Eample 6 Find f () if f() = ln. Since f( ) ln if 0 ln( ) if 0 it follows that f( ) 1 if 0 1 1 ( 1) if 0 Thus, f () = 1/ for all 0.
DERIVATIVES OF LOG FUNCTIONS Equation 4 The result of Eample 6 is worth remembering: d d ln 1
LOGARITHMIC DIFFERENTIATION The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simplified by taking logarithms. The method used in the following eample is called logarithmic differentiation.
LOGARITHMIC DIFFERENTIATION Eample 7 Differentiate y 3/ 4 2 (3 2) 5 1 We take logarithms of both sides of the equation and use the Laws of Logarithms to simplify: 3 1 2 ln y 4ln 2ln( 1) 5ln(3 2)
LOGARITHMIC DIFFERENTIATION Eample 7 Differentiating implicitly with respect to gives: 1 dy 3 1 1 2 3 5 y d 1 2 4 2 3 2 Solving for dy / d, we get: dy 3 15 y d 2 4 1 3 2
LOGARITHMIC DIFFERENTIATION Eample 7 Since we have an eplicit epression for y, we can substitute and write: 3/ 4 2 dy 1 3 15 5 2 d (3 2) 4 1 3 2
STEPS IN LOGARITHMIC DIFFERENTIATION 1. Take natural logarithms of both sides of an equation y = f() and use the Laws of Logarithms to simplify. 2. Differentiate implicitly with respect to. 3. Solve the resulting equation for y.
LOGARITHMIC DIFFERENTIATION If f() < 0 for some values of, then ln f() is not defined. However, we can write y = f() and use Equation 4. We illustrate this procedure by proving the general version of the Power Rule as promised in Section 3.1.
THE POWER RULE PROOF If n is any real number and f() = n, then f '( ) n n 1 Let y = n and use logarithmic differentiation: Thus, Hence, y' n ln y ln nln 0 y y ' n y n 1 n n n n
LOGARITHMIC DIFFERENTIATION You should distinguish carefully between: The Power Rule [( n ) = n n-1 ], where the base is variable and the eponent is constant The rule for differentiating eponential functions [(a ) =a ln a], where the base is constant and the eponent is variable
LOGARITHMIC DIFFERENTIATION In general, there are four cases for eponents and bases: d 1. ( a b ) 0 a and b are constants d d b b 1 f b f f 2. ( ) ( ) '( ) d d a a a g d g ( ) g ( ) 3. (ln ) '( ) g( ) 4. Tofind ( d / d[ f ( )],logarithmic differentiation can be used, as in the net eample.
LOGARITHMIC DIFFERENTIATION E. g. 8 Solution 1 Differentiate. y Using logarithmic differentiation, we have: ln y ln ln y ' 1 1 (ln ) y 2 1 ln 2 ln y ' y 2 2
LOGARITHMIC DIFFERENTIATION E. g. 8 Solution 2 ln Another method is to write. ( e ) d d d d ln ( ) ( e ) ln e d d 2 ln 2 ( ln )
THE NUMBER e AS A LIMIT We have shown that, if f() = ln, then f () = 1/. Thus, f (1) = 1. Now, we use this fact to epress the number e as a limit.
THE NUMBER e AS A LIMIT From the definition of a derivative as a limit, we have: f f (1 h) f (1) f (1 ) f (1) '(1) lim lim h 0 h 0 ln(1 ) ln1 1 lim lim ln(1 ) 0 0 1 lim ln(1 ) 0
THE NUMBER e AS A LIMIT As f (1) = 1, we have limln(1 ) 1 0 Formula 5 1 Then, by Theorem 8 in Section 2.5 and the continuity of the eponential function, we have: 1/ lim ln(1 ) 1 0 e e e e lim lim(1 ) 1 ln(1 ) 1 0 0 e lim(1 0 1 )
THE NUMBER e AS A LIMIT Formula 5 is illustrated by the graph of the function y = (1 + ) 1/ here and a table of values for small values of.
THE NUMBER e AS A LIMIT This illustrates the fact that, correct to seven decimal places, e 2.7182818
THE NUMBER e AS A LIMIT Formula 6 If we put n = 1/ in Formula 5, then n as 0 +. So, an alternative epression for e is: e lim 1 n 1 n n