can anyone help me finding first and second derivatives of these functions?
1) y=ln (x^2 +1)-x
2) y=(1+x^2)^(1/2)
3) y=tan x- x i got up to 2 sec^2xtanx
4) y=ln (x^2+1)
thanks lots
#1 ...
$\displaystyle
\frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx}
$
#2 ... chain rule
$\displaystyle
\frac{d}{dx} (u)^{\frac{1}{2}} = \frac{1}{2}(u)^{-\frac{1}{2}} \cdot \frac{du}{dx}
$
# 3 ...
$\displaystyle
\frac{d}{dx} (\tan{x} - x) = \sec^2{x} - 1 = \tan^2{x}
$
#4 ... use the rule given for #1.