How Do You Differentiate (sqrt(x))^x?

ssmh
Messages
1
Reaction score
0
I can't figure out
Derivative of (sq. rt of x)^x

Can anyone help me?
 
Physics news on Phys.org
Welcome to PF =]

Use the chain rule, this is simply a function within another function. the outer function just happens to be an exponential. \frac{d}{dx}f^x=f^x ln(f) (if I remember correctly)
 
if this is what you mean :

find f(x) prime, where f(x) = sqrt(x)^x

Before I help you, you need to help me. Tell me what concept you have learned that could
apply to this problem
 
djeitnstine said:
Welcome to PF =]

Use the chain rule, this is simply a function within another function. the outer function just happens to be an exponential. \frac{d}{dx}f^x=f^x ln(f) (if I remember correctly)

No, that's incorrect.
 
Here's how I did it.

\frac{d}{dx}(xx/2)

\frac{d}{dx}e(x/2)ln(x)

(1/2)(1 + ln(x))e(xln(x))/2

\frac{1 + ln(x)}{2} * x(x/2)
 
looks good to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top