Inverse Functions. Review from Last Time: The Derivative of y = ln x. [ln. Last time we saw that

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1 Inverse Functions Review from Last Time: The Derivative of y = ln Last time we saw that THEOREM The natural log function is ifferentiable an More generally, the chain rule version is ln ) =. ln u) = u u. EXAMPLE Determine D ln2 + )) 5. SOLUTION. Use the chain rule. D ln2 + )) 5 = D u 5 ) = 5u 4 u = 5ln2 + )) ln2 + )4 2 =. 2 + An Important Special Case. Since we can only take logs of positive numbers, often times we use the log of an absolute value, e.g., ln. We can fin the erivatives of such epressions as follows. D ln if > 0, D ln = D ln ) if < 0 = if > 0, ) = if < 0 = if = 0 In other wors, we get the same rule as without the absolute value: THEOREM For = 0, D ln ) = The chain rule version when u is a function of is ln u ) = u u. EXAMPLE Here s one that involves a number of log properties: D t ln e t cos t t 2 = D t ln e t + ln cos t ln t = D t t + ln cos t 2 ln t2 + = + 2t sin t) cos t 2t 2 + ) = tan t t t 2 +

2 math 30 logarithmic ifferentiation 2 YOU TRY IT 22.. Try fining these erivatives. Use log rules to simplify the functions before taking the erivative. a) D ln 6 3 sin b) D 6 3 ln sin ifferent) c) D ln ) D t lnt et) ) e) D ln f ) D s ln 5 ln s ) answer to you try it 22.. Use Theorem an log properties, then simplify. a) D ln 6 3 sin = D ln ln sin = 3 + cot b) 83 ln sin cot c) Dt e t ln t = e t e t ln t + e) D ln = D 3 ln = f ) Ds ln 5 ln s ) = Ds ln s ln 5 = ln In part f), remember that ln 5 is just a constant. + cot b) 83 D ln = D ln 4 ln 2 + = ) Dt lnt et ) ) = + t s

3 math 30 logarithmic ifferentiation 3 The Derivative of Other Eponential Functions: y = b When we first etermine the erivative of e we were looking at general eponential functions of the form y = f ) = b. Remember that we picke out e to be the e h number so that lim = which then mae computing the erivative of e h 0 h very easy. But what about all the other eponential functions of the form y = b? Is there a way to etermine their erivatives? Inee, there is! An it turns out to be pretty easy to o. Metho : This first evelopment of the erivative of y = b here is ifferent than in your tet. We will use implicit ifferentiation. Suppose that b > 0 an y = f ) = b. Our goal is to fin y = f ) = b ). Start with Take the logs of both sies Simplify Take the erivative Solve Substitute back for y y = b ln y = ln b ln y = constant {}}{ ln b ln y) = ln b y y = ln b y = y ln b y = b ln b That was easy. So we have THEOREM Derivative of b ). If b > 0, then for all, b ) = b ln b. The chain rule version when u is a function of is bu ) = b u ln b u. Metho 2: We coul also etermine the erivative of y = b using inverse functions. Because e an ln are inverse functions, one unoes the other, so if we apply both of them in succession, we en up with the original input. lne ) = an e ln) =. In particular, b = e ln b = e ln b where we use a log property at the last step to bring the power out front. So b is just an eponential function of the form e k where k = ln b is a constant. So b ) = e ln b) = ln be ln b = ln b)b = b ln b. 22.) We get the same result as above. This is the great avantage of working with logs: Logs turn proucts into sums an powers into proucts. These simplify many calculations.

4 math 30 logarithmic ifferentiation 4 EXAMPLE Fin the erivatives of the following functions. a) y = 2 b) z = 53 t/0 ) c) y = 5 t3 sin t ) y = 4 tan4) Solution. Using Theorem , a) 2 ) = 2 ln 2. b) This time we use the chain rule: ) {}}{ 53 t/0 ) = 53 t/0 ) ln 3 t 0 = 3 2 3t/0 ) ln 3. t 0 t ) c) Use the chain rule in combination with Theorem , ) Use the prouct rule: u t t 3 sin t) {}}{ 5 t 3 sin t) ) {}}{ = 5 t3 sin t ln 5 3t 2 sin t t 3 cos t). t 4 tan4)) = 4 ln 4 tan4) + 4 sec 2 4) 4 = 4 ln 4 tan4) + 4 sec 2 4)). Logarithmic Differentiation There are still types of functions that we have not trie to ifferentiate yet. Sometimes we can make use of our eisting techniques an clever algebra to fin erivatives of very complicate functions. Logarithmic ifferentiation refers to the process of first taking the natural log of a function y = f ), then solving for the erivative y. On the surface of it, it woul seem that logs woul only make a complicate function more complicate. But remember that logs turn powers into proucts an proucts into sums. That s the key. Here s a neat problem to illustrate the iea. EXAMPLE Use the chain rule an implicit ifferentiation along with logs to fin the erivative of y = f ) =. Solution. We begin by taking the natural log of both sies an simplifying using log properties. ln y = ln Powers = ln. Remember we want to fin y, so take the erivative of both sies implicitly on the left). ln y) = ln ) y y = ln + = ln) + y Solve = yln) + y Substitute = ln) + In other wors, we have shown that Here are a couple more. ) = ln) +. Neat! Easy!

5 math 30 logarithmic ifferentiation 5 EXAMPLE Fin the erivative of y = + 2 ) tan. Solution. Take the natural log of both sies an simplify using log properties. ln y = ln + 2 ) tan Powers = tan ln + 2 ). Take the erivative of both sies implicitly on the left) an solve for y. ) ln y) = tan ln + 2 ) y y = sec2 ln ) + tan + 2 y Solve = y sec 2 ln tan ) y Substitute = ln + 2 ) tan sec 2 ln + 2 ) + So ln + 2 ) tan ) = ln + 2 ) tan sec 2 ln + 2 ) + EXAMPLE Fin the erivative of y = ln ) 3. 2 tan tan + 2. Not ba! Solution. Be careful. This function is NOT the same as ln 3 ) which woul equal 3 ln. Instea, take the natural log of both sies an simplify using log properties. ln y = lnln ) 3 Powers = 3 lnln ). Take the erivative of both sies implicitly on the left) an solve for y. y y = 32 lnln ) + 3 ln 3 2 lnln ) + 3 ln y y Solve = y y Substitute = ln ) lnln ) + 3 ln Do you see the ifference when compare to ln 3 ) Logs can also be use to simplify proucts an quotients. EXAMPLE Fin the erivative of y = 2 ) Solution. Use logarithmic ifferentiation to avoi a complicate quotient rule erivative Take the natural log of both sies an then simplify using log properties. 2 ) 5 ) + 2 ln y = ln Log Prop = ln 2 ) 5 + ln + 2 ) /2 ln 4 + 4) Log Prop = 5 ln 2 ) + 2 ln + 2 ) ln 4 + 4). Take the erivative of both sies an solve for y.

6 math 30 logarithmic ifferentiation 6 y y = y Solve = y y Substitute = 2 ) That woul have been a real mess to o with the quotient rule which woul also require the prouct rule an the chain rule). Problems The following questions will be on the lab tomorrow or are future WebWorK problems. Get a hea start.. Fin the erivatives of the following functions. Use logarithmic ifferentiation where helpful. a) y = sin ) b) y = sin c) sin ) sin ) arcsin ) 2 e) + ) 2. Fin the erivatives of these functions using the erivative formula for a general eponential function that we evelope before Eam II. See Theorem 3.8 on page 94). a) 5 6 b) 2 cot c) π + π ) 4 4 e) For which values of oes 4 4 have a horizontal tangent? Answers.. a) ln y = lnsin ) = lnsin ) y y sin ) lnsin ) + cot ). cos = lnsin ) + sin b) ln y = ln sin = sin ln y y = cos ln + sin ) y = sin y = cos ln + sin ). c) ln y = lnsin ) sin = sin lnsin ) y y cos = cos lnsin ) + sin ) sin y = sin ) sin cos lnsin ) +. ) ln y = lnarcsin ) 2 = 2 lnarcsin ) y y = 2 lnarcsin ) + 2 arcsin 2 ) y = arcsin )2 2 lnarcsin ) + 2 arcsin ). 2 e) ln y = ln + ) = ln + ) y y = ln + ) + ) + 2 y y + = ln ) + ) ln + ) ) + y y = ln. + ) + ) 2. a) 5 6 = 5 6 ln 6 = 5 ln 66 ). b) 2 cot = 2 ln 2 cot 2 csc 2 = 2 ln 2 cot csc 2. + ) y = c) π + π = π π + π ln π ) 4 4 = ln 4 = ln 4 e) From the previous part, the slope is 0 when ln 4 = 0. Therefore = 0 or = 4 ln 4.