by TSN79
Tags: differentiating
TSN79 is offline
Oct6-05, 12:10 PM
P: 355
I'm having some trouble differentiating [itex] x^{\sqrt x } [/itex]. I know that the derivative of [itex] x^{\sqrt x } [/itex] probably begins with [itex] x^{\sqrt x } \cdot \ln (x) \cdot \frac{1}{{2\sqrt x }} [/itex] but once the base is also x then there is probably more to it than that. Anyone?
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
siddharth is offline
Oct6-05, 12:26 PM
HW Helper
PF Gold
siddharth's Avatar
P: 1,198
You have [tex] y=x^{\sqrt x } [/tex]
So, [tex] \ln y = (\sqrt x)(\ln x) [/tex]
Then differentiate both sides with respect to x and substitute the value of y.
TSN79 is offline
Oct6-05, 12:50 PM
P: 355
Ah, yes, thanks siddharth!

Register to reply

Related Discussions
Differentiating x^x Precalculus Mathematics Homework 10
Differentiating this Calculus & Beyond Homework 7
Differentiating Calculus & Beyond Homework 1
Help Differentiating Calculus & Beyond Homework 4
Differentiating Ln Calculus & Beyond Homework 7