What is the derivative of #x^(3/x)#?

1 Answer
Luke Phillips
Jun 4, 2017

#"d"/("d"x) x^(3/x) = (ln(3/x)-3)*x^(3/x)#

Explanation:

Write #x^(3/x) = exp(-xln(x/3))#.

Then, by the chain rule,
#"d"/("d"x) x^(3/x) = exp(-xln(x/3))*"d"/("d"x)(-xln(x/3))#,

#"d"/("d"x) x^(3/x) = exp(-xln(x/3))*(-ln(x/3)-x*1/(x/3))#,

#"d"/("d"x) x^(3/x) = exp(-xln(x/3))*(ln(3/x)-3)#,

#"d"/("d"x) x^(3/x) = (ln(3/x)-3)*x^(3/x)#.

Related questions
See all questions in Differentiating Exponential Functions with Other Bases
Impact of this question
1366 views around the world
You can reuse this answer
Creative Commons License