How do you find the derivative of $x^{\frac{3}{x}}$ $x^(3/x)$ ?

Question

1272 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-08T22:40:03

$= 3 x^{\frac{3}{x} - 2} (1 - ln x)$ $=3 x^(3/x - 2) (1-ln x)$

use logs cos that's ugly!

$y = x^{\frac{3}{x}}$ $y = x^(3/x)$

$ln y = {ln x}^{\frac{3}{x}} = \frac{3}{x} ln x$ $ln y = ln x^(3/x) = 3/x ln x$

then implicit diff.... plus product rule on the RHS
$1/y \ y' = - 3/x^2 ln x + 3/x* 1/x$

$1/(x^(3/x)) \ y' = 3/x^2 (1-ln x)$

$y' =(x^(3/x)) 3/x^2 (1-ln x)$

$=3 (x^(3/x)) x^(-2) (1-ln x)$

$=3 x^(3/x - 2) (1-ln x)$

How do you find the derivative of x^(3/x)x3xx^(3/x)?