Asked by Lindsay
Find the differential dy of the given function.
y=sec^2x/(x^2+1)
My answer was [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2
My friend (who's a lot smarter than me) got something wayyy different, so I'm not sure if I'm doing this wrong...?
y=sec^2x/(x^2+1)
My answer was [tanx(x^2+1)-2sec^2x]dx/(x^2+1)^2
My friend (who's a lot smarter than me) got something wayyy different, so I'm not sure if I'm doing this wrong...?
Answers
Answered by
Damon
[(x^2+1)d(sec^2x)-sec^2xd(x^2+1)]/(x^2+1)^2
[(x^2+1)2secxdsecx-sec^2x(2xdx)]/(x^2+1)^2
[2(x^2+1)sec^2xtanxdx-2xsec^2xdx]/(x+1)^2
2 sec^2 x dx [(x^2+1)tan x - x ]/ (x+1)^2
[(x^2+1)2secxdsecx-sec^2x(2xdx)]/(x^2+1)^2
[2(x^2+1)sec^2xtanxdx-2xsec^2xdx]/(x+1)^2
2 sec^2 x dx [(x^2+1)tan x - x ]/ (x+1)^2
Answered by
Lindsay
Wait, I'm not understanding how you get the tanx -x part...
Answered by
Lindsay
Hold on...I think I just managed to kind of figure it out!