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Unformatted text preview: Notes on the Differentiation of the Natural Logarithm Function This lesson is entirely about differentiating function involving the natural logarithm function. The homework problems make use of the rules for differentiation that were discussed in Calculus I, including the Chain Rule. In addition you need to remember some of the applications of differentiation form Calculus I. AS you do the homework, concentrate on your presentation. The examples given below illustrate what I am expecting in your presentation. You will notice that many of the examples include extra steps, that is, steps showing how to do the differentiation (e.g., the Product Rule or Quotient Rule) or simplify the answer algebraically. You dont have to include every one of these in your presentation. I include them here to help you remember them. From the definition of the natural logarithm function it follows that 1 ln x dt x t = for all0 x < < . As shown previously, this means that 1 ln d x dx x = for0 x < < . By the Chain Rule, if f is a differentiable function and positive on an interval I , then ( 29 ( 29 ( 29 ln f x d f x dx f x = for all values of x in I . Now if x < < , then0 x <  < and ( 29 ( 29 ( 29 1 1 1 ln 1 d d x x dx x dx x x = =  = . Therefore the following formula holds: 1 ln d x dx x = , x . Again by the Chain Rule, if f is differentiable function and ( 29 f x an interval I , then ( 29 ( 29 ( 29 ln f x d f x dx f x = for all x in I . Example 1 : Evaluate ( 29 2 ln 2 d x dx . Solution : ( 29 ( 29 2 2 2 1 ln 2 2 2 d d x x dx x dx = [ ] 2 2 2 2 1 1 2 2 0 2 2 2 2 d d x x x x dx dx x x = = =  . ( In this example, the differentiation is only valid when 2 2 x that is 2 2 x < < . ) In the next example the actual computations are the very similar, but where the differentiation is valid is different because the absolute value is introduced. Example 2 : Evaluate 2 ln 2 d x dx . 1 Solution : ( 29 2 2 2 1 ln 2 2 2 d d x x dx x dx = [ ] 2 2 2 2 1 1 2 2 0 2 2 2 2 d d x x x x dx dx x x = = =  . ( In this example, the differentiation is valid when 2 2 x that is 2 x  and 2 x . ) Example 3 : Let ( 29 3 ln 1 ln t t f t t = + be define by. Find ( 29 f t . Solution : ( 29 3 ln 1 ln d t t f t dt t = + ( 29 ( 29 ( 29 ( 29 ( 29 2 3 ln 1 ln 1 ln 3 ln 1 ln d d t t t t t t dt dt t + + = + ( 29 ( 29 ( 29 2 1 1 3 ln 1 ln 3 1 ln t t t t t t  + + = + ( 29 ( 29 ( 29 ( 29 2 3 ln 1 ln 3 1 1 ln t t t t t t + = + ( 29 2 1 3 ln 1 ln t t t t = + ....
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 Spring '11
 BUEKMAN
 Calculus

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