View Single Post
dextercioby
#7
Mar30-07, 06:24 AM
Sci Advisor
HW Helper
P: 11,893
[tex] y=\arctan x [/tex]

[tex] \tan y =x [/tex]

[tex] \frac{d}{dx}\tan y= \frac{1}{\cos^{2}y}\frac{dy}{dx}=1 [/tex]

[tex] \frac{dy}{dx} =\cos^{2} y=\left(\frac{1}{\sqrt{1+x^{2}}}\right)^{2}=\frac{1}{1+x^{2}} [/tex]