Calculus- Early Transcendentals, 2021a

152 Derivatives
Solution. We take ln of both sides:
Using log properties we have:
Differentiating implicitly gives:
Solving for y ′ gives:
Replace y = x x gives:
lny = lnx x .
lny = xlnx.
y ′
y =(1)lnx + x1 x .
y ′
= lnx + 1.
y
y ′ = y(1 + lnx).
y ′ = x x (1 + lnx).
Another method to find this derivative is as follows:
d
dx xx = d dx exlnx
( ) d
=
dx xlnx e xlnx
= (x 1 x + lnx)xx
= (1 + lnx)x x ♣
In fact, logarithmic differentiation can be used on more complicated products and quotients (not just
when dealing with functions to the power of functions).
Example 4.47: Logarithmic Differentiation
Differentiate (assuming x > 0):
y = (x + 2)3 (2x + 1) 9
x 8 (3x + 1) 4 .
Solution. Using product and quotient rules for this problem is a complete nightmare! Let’s apply logarithmic
differentiation instead. Take ln of both sides:
( (x + 2) 3 (2x + 1) 9 )
lny = ln
x 8 (3x + 1) 4 .
Applying log properties:
lny = ln ( (x + 2) 3 (2x + 1) 9) − ln ( x 8 (3x + 1) 4) .
lny = ln ( (x + 2) 3) + ln ( (2x + 1) 9) − [ ln ( x 8) + ln ( (3x + 1) 4)] .