## anonymous 5 years ago what is the derivative of (lnx^2)/(x+1)^(1/2)

1. amistre64

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

2. amistre64

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

3. amistre64

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

4. amistre64

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

5. amistre64

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

6. amistre64

b = (x+1)^(1/2) ; b' = 1/[2(x+1)^(1/2)] t = ln(x) ; t' = 1/x

7. amistre64

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

8. amistre64

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

9. amistre64

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

10. anonymous

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.