Differentiation ln [radical ((x^2) + 4))]

[skyblue]

ln [radical ((x^2) + 4))] it is all under the radical.

i did U'/U

U'=2x
U=(x^2) +4

so 2x / (x^2)+4

now is the answer: x / (x^2)+4

what happened to the 2 then from (2x)/(x^2)+4? can someone please help me. thank you.

 

[skeeter]

\(\displaystyle \L y = \ln{\sqrt{x^2+4}}\)

\(\displaystyle \L y = \ln{(x^2+4)^{\frac{1}{2}}}\)

\(\displaystyle \L y = \frac{1}{2} \ln{(x^2 + 4)}\)

\(\displaystyle \L y' = \frac{1}{2} \cdot \frac{2x}{x^2 + 4}\)

\(\displaystyle \L y' = \frac{x}{x^2 + 4}\)