209k views
4 votes
Using (ln x)' = 1/x, differentiate the function f(x)=x^x

User YahyaE
by
7.6k points

1 Answer

4 votes

You have the following function:

f(x) = xˣ

In order to differentiate the previous function, you proceed as follow:

apply ln to the function:

ln f(x) = ln (xˣ)

use the property of logarithm ln aⁿ = n(lna)

ln f(x) = x (ln x)

differentiate the previous equation

[ln f(x)]' = ln x + x(1/x) = ln x + 1

[ln f(x)]' = ln x + 1

use [ln f(x)]' = f'(x)/f(x)

f'(x)/f(x) = ln x + 1

f'(x) = f(x)[ln x + 1]

replace f(x) = xˣ

f'(x) = xˣ[ln x + 1]

User Baleb
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories