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!
    
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