Math 127  Fall 05   Notes on  Logarithmic Differentiation                                                             p.1 of  3

     Sometimes we encounter problems for which all our shortcuts learned in Chapter 3 do not help.

Example:   Suppose  y = f(x) = x^x  for  x > 0.

Here the power rule does not apply because the exponent  x  is NOT a constant and the rule for

exponentials does not apply because the base  x  is NOT a constant.  Furthermore, none of the

other rules seems to apply either so it seems there is no shortcut for this problem which means

that we will have to do this problem by referring back to the original definition of the derivative

as a limit of slopes of secant lines.  However, this approach leads to rather unpleasant limits which

you may not be able to figure out.  

     Fortunately, there is another way which leads to a method called LOGARITHMIC

DIFFERENTIATION which can be used not only for this problem but much more generally for

a variety of problems where none of our other shortcuts apply or when these shortcuts lead to very

messy algebra.  Furthermore, this method can be used to prove the power rule for all exponents  n

(our previous proof only worked when  n  was a positive integer), an alternate (and much simpler)

proof of the product rule and also a proof of the quotient rule which we earlier stated could be

proved by algebra similar to (but much more complicated than) that in the original proof of the

product rule and therefore omitted that proof from our lectures.

     Solution: (of our original problem  y = f(x) = x^x.) 

                      Step I:    Take  ln of both sides:   ln(y) = ln(x^x).

                      Step II:   Simplify the right side using properties of logs:   ln(y) = xln(x).

                      Step III:  Differentiate both sides using the chain rule on the left side and the product

                                   rule on the right side:  (1/y) dy/dx  =  x(1/x) + (1)ln(x) = 1 + ln(x)

                      Step IV:  Solve for the desired  dy/dx  by multiplying both sides by  y  and then rewriting 

                                  y  in terms of  x  as in the original problem:  dy/dx = y(1 + ln(x)) = x^x(1 + ln(x)).

Math 127  Fall 05   Notes on  Logarithmic Differentiation                                                             p.2 of  3

The steps used in the particular problem above are the steps used for the method in general

although the particular properties of logs used in this problem may be replaced by different

properties of logs in other problems.  We illustrate this by giving new proofs of the power rule,

product rule and quotient rule.

     Power Rule:  If   y = f(x) = xⁿ  where  n is a (constant) real number, then  y' = dy/dx = nx^n-1.

     Proof: (By logarithmic Differentiation):  Step I:  ln(y) = ln(xⁿ).  Step II:  ln(y) = n ln(x).

                StepIII:  (1/y) dy/dx = n (1/x).  Step IV:  dy/dx = y n (1/x) = xⁿ n (1/x) = n x^n-1.

     Remark:  This proof is valid for any real number  n,  whereas the original proof in lecture was

                     only valid when  n  is a positive integer.

     Product Rule:  If  y = f(x) g(x),  then  y' = dy/dx = f'(x) g(x)) + f(x) g'(x).

     Proof: (By Logarithmic Differentiation):  Step I:  ln(y) = ln[f(x) g(x)].  Step II:  ln(y) = ln[f(x)] + ln[g(x)].

                StepIII:  (1/y) dy/dx = [1/f(x)] f'(x) + [1/g(x)] g'(x).  Step IV:  dy/dx =

                y {[1/f(x)] f'(x) + [1/g(x)] g'(x)} = [f(x) g(x)] {[1/f(x)] f'(x) + [1/g(x)] g'(x)} = g(x) f'(x) + f(x) g'(x).

     Quotient Rule:  If  y = f(x)/g(x),  then  y' = dy/dx = [f'(x) g(x)) − f(x) g'(x)]/(g(x))².

      Proof: (By Logarithmic Differentiation):  Step I:  ln(y) = ln[f(x)/g(x)].  Step II:  ln(y) = ln[f(x)] − ln[g(x)].

                StepIII:  (1/y) dy/dx = [1/f(x)] f'(x) − [1/g(x)] g'(x).  Step IV:  dy/dx =

               y {[1/f(x)] f'(x) − [1/g(x)] g'(x)} = [f(x)/g(x)] {[1/f(x)] f'(x) − [1/g(x)] g'(x)} = 

              f'(x)/g(x) − f(x) g'(x)/(g(x))²  =  [f'(x) g(x) − f(x) g'(x)]/(g(x))².

     The method can also be used for problems which could be done by combinations of the original

 shortcut rules, but for which the algebra doing it that way becomes quite messy:

Math 127  Fall 05   Notes on  Logarithmic Differentiation  

                                                          p.2 of  3

Example:  If  y = f(x) = [e^x ln(x)]/(x² + 1), then  y' = dy/dx   could be computed by the quotient rule, in

combination with the product rule, which would be needed to compute the derivative of the

numerator.  The calculation would go something like this:

     y' = dy/dx = {[e^x ln(x)]' (x² + 1) − [e^x ln(x)] (x² + 1)'}/(x² + 1)²  =  {[(e^x) ln(x) + e^x (1/x)] (x² + 1) −

             [e^x ln(x)] (2x)}/(x² + 1)²  =  {e^x( ln(x) + (1/x))(x² + 1) − e^x ln(x) (2x)}/(x² + 1)²  =

             {e^x[ ln(x) (x² + 1)  + (x² + 1)/x − ln(x) (2x)]}/(x² + 1)²  =  {e^x [(x² − 2x +1) ln(x) + (x + (1/x))]/(x² + 1)².

     If we use Logarithmic Differentiation on this same problem the calculation goes like this:

     Step I:  ln(y) = ln{[e^x ln(x)]/(x² + 1)}.  Step II:  ln(y) = ln(e^x) + ln(ln(x)) − ln((x² + 1)). 

     StepIII:  (1/y) dy/dx = 1 + (1/ln(x)) (1/x) − (1/(x² + 1)) (2x) = 1 + 1/(x ln(x)) + 2x/(x² + 1).

     Step IV:  dy/dx = y[1 + 1/(x ln(x)) + 2x/(x² + 1)] = {[e^x ln(x)]/(x² + 1)}[ 1 + 1/(x ln(x)) + 2x/(x² + 1)].

     After some algebra it can be shown that these two formulas for  dy/dx, namely the one in the

     original calculation and the one by logarithmic Differentiation, are equal but either answer is an

     acceptable calculation of  dy/dx  and it seems to me that the algebra in the calculation by Logarithmic

     Differentiation is considerably simpler that that in the original method.