Question

Differentiate y = x^1/2 ln x

Answer 1

take ln for both sides and do it implicitly.

Answer 2

dy/dx = x^1/2 (1/x) + 1/2x^-1/2 (lnx) right?

Answer 3

omg i cant really read what you wrote. :P Sorry.

Answer 4

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s06_qp_3.pdf The ques is no.8 i)

Answer 5

agreed.

just simplify it.

Answer 6

So, is mine right?

Answer 7

thats right!\[\dfrac{dy}{dx} = x^{1/2} (1/x) + 1/2x^{-1/2} (\ln x) \]

set that equal to \(0\) and solve \(x\)

Answer 8

How do i simplify it. D:

Answer 9

not to butt in, but is this
\[y=\sqrt x\ln(x)\]?

Answer 10

no prob. yes it is

Answer 11

then i would say rational exponents are not your friends here
go with
\[\frac{\ln(x)}{2\sqrt{x}}+\frac{\sqrt{x}}{x}\] which is easy to add

Answer 12

especially since \(\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\)

Answer 13

how do i simplify it ffrom there

Answer 14

@ganeshie8

Answer 15

set that equal to 0 and solve x

Answer 16

can't solve it D:

Answer 17

show ur attempt

Answer 18

-sqrt x /x = 1/2 x^-1/2 lnx
dont know how to continue anymore.

Answer 19

esp on the right side, how should i simplify it?

Answer 20

you're almost there !
simply cancel out x^(-1/2) both sides

Answer 21

Notice :
\[\sqrt{x}/x = x^{1/2-1} = x^{-1/2}\]

Answer 22

but when you move it to the other side would be a negative alrd. D:

Answer 23

yes

Answer 24

Hey i got it alrd!!! thanks! :))

Answer 25

dy/dx = x^1/2 (1/x) + 1/2x^-1/2 (lnx) = 0

x^(-1/2) + 1/2 x^(-1/2) lnx = 0

x^(-1/2) [2 + lnx] = 0

2 + lnx = 0
lnx = -2

x = e^(-2) = 1/e^2

Answer 26

exsctly.

Answer 27

good good