Question

derivative of exponents

Answer 1

\[y=\sqrt{x ^{x}}\]

Answer 2

would this be equivalent to\[y=x ^{\frac{ 1 }{ 2 }x}\]

Answer 3

then I could take the ln and use derivative\[\ln y=\frac{ 1 }{ 2 }x\]

Answer 4

you just use chain rule

Answer 5

can you explain?

Answer 6

f prime of g(x)

Answer 7

square root of x^n is two functions

Answer 8

the base of square root is x^1/2

Answer 9

derivative of x^1/2 and plug x^n in times the derivative of x^n

Answer 10

Im not sure how to do that I see what you're saying but Im stuck

Answer 11

would it be \[y=\frac{ 1 }{ 2\sqrt{x} }\sqrt{x}x ^{\frac{ 1 }{ 2 }-1}\]

Answer 12

@aznandmaths

Answer 13

you could use
\[ x = e^{\ln(x)} \]
and write the problem as
\[ \sqrt{x ^{x}} = x^\frac{x}{2} = e^\frac{x\ln(x)}{2}

\]

Answer 14

could I still take the ln though as you have it in the second step so that it is \[lny=\frac{ x }{ 2 }lnx\]

Answer 15

yes, thats correct

now you'll need product rule for right side

Answer 16

so it would be \[\frac{ y' }{ y }=1/2lnx+1/2\]

Answer 17

correct

Answer 18

rewrite y as sqrt (x^x)

Answer 19

\[y'=\frac{ 1 }{ 2 }\sqrt{x ^{x}}(lnx+1)\]

Answer 20

thank you so much

Answer 21

correct! and welcome ^_^