On-line Math 21

3.3.4  Derivatives of other Exponential and Logarithmic functions

  Derivatives of exponential functions

Any other exponential function, that is, any function a^x where a is not e - this shows my prejudice in favor of e when dealing with exponentials - can be interpreted in terms of e^x , and so its derivatives can as well.

If f(x) = a^x , then, since

a = eln(a),

f(x) = ax = ( eln(a)) x = eln(a) x = ex ln(a).

No derivatives, yet. But, to compute the derivative of f(x) , you use the chain rule and the last version above for f(x) :

f¢(x) = ( ax) ¢ = ( ex ln(a)) ¢ = ex ln(a)(x ln(a))¢ = ln(a) ax.

So, this is no big deal. But it does show why e is so useful for calculus; had we stuck with our biological preference for base 10, we'd have to remember that

( 10x) ¢ = 2.3026 10x.

  Logarithms.

For other bases, the formula is simply altered by the rule

logb(x) =

lnx lnb

,

( logb(x)) ¢ = 1 x lnb .

This is analogous to the formula for ( b^x) ¢ = b^xlnb , which we just derived.

Example 1

d dx ln æ ç ç ç è x+1   ___ Öx-2 ö ÷ ÷ ÷ ø = d dx æ è ln( x+1) -ln æ è   ___ Öx-2   ö ø ö ø = d dx æ ç è ln( x+1) - 1 2 ln( x-2) ö ÷ ø = 1 x+1 - 1 2(x-2) .

Actually, the technique used here, using the rules of logarithms to simplify the expression before the derivative is computed, is sometimes useful. It is referred to as logarithmic differentiation, and has the advantage of replacing hard derivative rules (like the quotient rule) with easier ones, like the sum rule.

The general idea is that, if y = f(x) is a ``messy'' expression, then perhaps ln(f(x)) can be simplified, then differentiated,

d dx ln(f(x)) = 1 f(x) df dx .

But, the assumption is that the left-hand side of this is easer to compute than the right, so you find f¢(x) by computing the left-hand side, and:

f(x) d dx ln(f(x)) = df dx .

Example 2 If

f(x) = (x2+3)4e3x (x-3)(2x+5)2 ,

find f¢(x) .

Solution

Exercise 1 Find

æ ç ç ç è Ö   x2+3x+1   ( x2+3) 2 (3x+2)4 ö ÷ ÷ ÷ ø ¢

using implicit differentiation:

Answer

Example 3

( xsinx) ¢ = ( ( elnx) sinx) ¢ = ( elnx sinx) ¢ = elnx sinx( sinx x +lnx cosx) = xsinx( sinx x +lnx cosx).

Exercise 2 Find

( xxx) ¢ =

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