Question

what is the derivative of (lnx^2)/(x+1)^(1/2)

Answer 1

\[D_x(\frac{\ln(x)}{(x+1)^2})\]

Answer 2

this thing? and whered my fraction bar end up :)

Answer 3

use quotient rule; or bring the botom up with a negative exponent and use product rule

Answer 4

murmuring..... ^(1/2) lol

Answer 5

bt'-b't
f'(x) = -----
b^2

Answer 6

b = (x+1)^(1/2) ; b' = 1/[2(x+1)^(1/2)]

t = ln(x) ; t' = 1/x

Answer 7

b^2 = (x+1) :) that one was easy lol

Answer 8

\[\frac{\frac{\sqrt{x+1}}{x} - \frac{\ln(x)}{2 \sqrt{x+1}}}{(x+1)}\] simplify as needed

Answer 9

2x + 2 - x ln(x)
---------------
2x(x+1) sqrt(x+1)

Answer 10

amistre64,

In the following equation, Ln is spelled Log.
\[\int\limits \frac{\frac{\sqrt{1+x}}{x}-\frac{\text{Log}[x]}{2 \sqrt{1+x}}}{1+x} \, dx = \frac{\text{Log}[x]}{\sqrt{1+x}} \]
It looks to me like the Ln of x was raised to the second power in the original problem statement above. Perhaps by eyes are failing me in my old age.