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Unit 10: Logarithmic Differentiation
Notes
g(x)
Notes The given function contains a term of the form ( f(x)) , with f(x) = sin x and g(x) = cos x.
g(x)
Hence we use either the equation ( f(x)) = e g(x) ln f(x) as in Alternate 1 or logarithmic
differentiation as in Alternate 2. Again, in the answer don’t forget to replace e g(x) ln f(x) by
g(x)
( f(x)) , or y by the expression of the given function.
Example: Find:
d (1 x )(2 ) x 2 (3 ) x 3
.
dx (4 ) x 4
Solution:
Let:
(1 x )(2 ) x 2 (3 ) x 3
y .
(4 ) x 4
Employing logarithmi c differenti ation we obtain :
ln y ln 1 ( ) x 2ln 2 ( ) x 3ln 3 ( ) x 4ln 4 ( x ),
1 dy 1 2 3 4
y dx 1 x 2 x 3 x 4 x ,
d (1 x )(2 ) x 2 (3 ) x 3 dy
dx (4 ) x 4 dx
1 2 3 4
y
1 x 2 x 3 x 4 x
(1 x )(2 ) x 2 (3 ) x 3 1 2 3 4 .
(4 ) x 4 1 x 2 x 3 x 4 x
g(x)
Notes Here we have a product and a quotient, but there’s no term of the form ( f(x)) , and
we still employ logarithmic differentiation, which therefore isn’t exclusive for the form
g(x)
( f(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.
Example: Differentiate y = (sec x) tan x in 2 ways:
1. Express it as natural exponential and then differentiate.
2. Use logarithmic differentiation.
Solution:
1. y = (sec x) tan x = e tan x ln sec x ,
y’ = e tan x ln sec x (sec x ln sec x + tan x (1/sec x) sec x tan x)
2
2
2
= (sec x) tan x (sec x ln sec x + tan x).
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