Examples of Logarithmic Differentiation
General Comments
Logarithmic differentiation makes things a lot nicer in many cases, but there are usually other methods that you could
use (if you’re willing to work through some messy differentiation). However, if you have a function that looks like a
f (x)
function raised to another function, i.e., y = [f1 (x)] 2 where f1 , f2 are not constant, then it’s the only method you can
use. Note: If either f1 or f2 were constant, then we’d have other rules that we could use: for example, to differentiate
2
y = 101−x we could use chain+exponential rules, and to differentiate y = (1 − x2 )10 we could use chain+power rules.
Examples
Example 1. Differentiate
cos(x)
y = (ln(x))
We first take ln of both sides and bring the cos(x) out of the exponent:
cos(x)
ln(y) = ln (ln(x))
= cos(x) ln (ln(x))
We then differentiate implicitly, using the product and chain rules on the RHS:
1 1
1 0
y = − sin(x) ln (ln(x)) + cos(x)
y
ln(x) x
Finally, multiply both sides by y, writing the RHS in terms of x:
cos(x)
cos(x)
y0 = − sin(x) ln (ln(x)) +
(ln(x))
x ln(x)
Example 2. Differentiate
y = xx
x
x
2
Note: This is read as x(x ) , not (xx )x (which is equal to x(x ) ).
We first take ln of both sides and bring xx out of the exponent:
x
ln(y) = ln xx
= xx ln(x)
Now at this point, we still have that xx term that we can’t use any of our usual rules on. But, we are allowed to take ln
of both sides a second time and then use properties of logs to split up the RHS:
ln(ln(y)) = ln (xx ln(x))
= ln(xx ) + ln(ln(x))
= x ln(x) + ln(ln(x))
Now we can differentiate implicitly (there are multiple chain rules plus a product rule being used here):
1 1 0
1
1 1
y = ln(x) + x +
ln(y) y
x ln(x) x
1
Finally, solve for y 0 and write the RHS in terms of x:
1
y0 = ln(x) + 1 +
y ln(y)
x ln(x)
x
x
1
= ln(x) + 1 +
xx ln xx
x ln(x)
2