Calculus Of One Real Variable – By Pheng Kim Ving Chapter 7: The Exponential And Logarithnic Functions – Section 7.4: Logarithmic Differentiation 7.4 Logarithmic Differentiation

 1. Differentiation Of The Function y = ( f (x)) g (x)

Let's differentiate the function y = ( f (x))g(x), where both the base and the exponent are functions of x. Since the variable
appears in both the base and the exponent, we can't apply either the power rule or the exponent rule to this function directly.

But first we define ( f (x))g(x) in terms of f (x) and g(x). For x where f (x) > 0, we define ( f (x))g(x) = eg(x) ln f (x). The expression
ln f (x) requires that f (x) > 0. That definition is similar to that of bx, which is bx = ex ln b. For x where f (x) = 0 and g(x) > 0,
we define (
f (x))g(x) = 0. Of course 0g(x) necessitates that g(x) > 0 for it to be defined. Method 2

Taking the natural logarithm of both sides of y = ( f (x))g(x) we have:

ln y = g(x) ln f (x),

differentiate both sides: Logarithmic Differentiation

In Method 1 we transform (  f (x))g(x) into the exponential function using the definition uv = ev ln u, to get ( f (x))g(x) =
eg(x) ln f (x). Then we differentiate eg(x) ln f (x) with respect to x utilizing the exponent rule. This is possible because the base e is a
constant. In the answer, we transform
eg(x) ln f (x) back to ( f (x))g(x).

In Method 2 we take the natural logarithm of both sides of the equation y = ( f (x))g(x), to obtain ln y = ln ( f (x))g(x) =
g(x) ln f (x). Then we differentiate implicitly both sides of the resulting equation ln y = g(x) ln f (x) with respect to x. Note
that (
d/dx) ln y = (1/y) dy/dx = y'/y, by the chain rule. Next we solve for y '. In the answer, we replace y by ( f (x))g(x),
since we should express
y ' in terms of x only, not of x and y. This technique is called logarithmic differentiation, because
it involves the taking of the natural logarithm and the differentiation of the resulting logarithmic equation. It allows us to
convert the differentiation of (
f (x))g(x) into the differentiation of a product.

Note that both Methods 1 and 2 yield the same answer.

Remark 1.1

In practice, you don't have to memorize the formula of the derivative of ( f (x))g(x). That's welcome, because it's not easy to
memorize it. Just remember one or both of the methods themselves.

Example 1.1

Differentiate y = (2x)sin x in 2 different ways.

Solution 1
y = (2x)sin x = esin x ln 2x, EOS

# Solution 2

ln y = ln (2x)sin x = sin x ln 2x, EOS

Of course in practice just use one method, unless you're asked to use both methods.

 2. Differentiation Of Expressions Containing ( f (x)) g (x)

Example 2.1

Find dy/dx if y = x3(sin x)cos x using 2 different methods.

Solution 1

y = x3(sin x)cos x = x3ecos x ln sin x, EOS

# Solution 2 Utilizing logarithmic differentiation we get:

ln y = ln x3 (sin x)cos x = 3 ln x + cos x ln sin x, EOS

The given function contains a term of the form ( f (x))g(x), with f (x) = sin x and g(x) = cos x. Hence we use either the
equation (
f (x))g(x) = eg(x) ln f (x) as in Solution 1 or logarithmic differentiation as in Solution 2. Again, in the answer don't forget to
replace
eg(x) ln f (x) by ( f (x))g(x), or y by the expression of the given function.

Again in practice just use one method, unless you're asked to use both methods.

 3. Differentiating Products And Quotients

Example 3.1

Find: # Solution Let: EOS

Here we have a product and a quotient, but there's no term of the form ( f (x))g(x), and we still employ logarithmic
differentiation, which therefore isn't exclusive for the form (
f (x))g(x). Of course we can use the product and quotient rules, but
doing so would be more complicated. Generally, logarithmic differentiation is advantageous when the products and/or
quotients are complicated. It enables us to convert the differentiation of a product and that of a quotient into that of a sum and
that of a difference respectively.

 Problems & Solutions

1.  Differentiate y = (sec x)tan x in 2 ways:
a.  Express it as natural exponential and then differentiate.
b.  Use logarithmic differentiation.

Solution

a.  y = (sec x)tan x = etan x ln sec x,

y' = etan x ln sec x (sec2 x ln sec x + tan x (1/sec x) sec x tan x)

= (sec x)tan x (sec2x ln sec x + tan2 x).

b.  ln y = ln (sec x)tan x = tan x ln sec x, y' = y(sec2 x ln sec x + tan2 x) = (sec x)tan x (sec2 x ln sec x + tan2 x). 2.  Find y' using logarithmic differentiation if y = xx/(x – 1)2.

Solution   a.  Which of these functions grows more rapidly for sufficiently large x?
b.  Differentiate them.

Solution   4.  Find: Solution

Let:  5.  Find an equation of the tangent line to the curve: at x = 0.

Solution Consequently the equation of the tangent line is y – 1 = 0(x – 0) or y = 1.