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Question Number 11011 by madscientist last updated on 06/Mar/17
if tan(xy)=x then (dy/dx)=
iftan(xy)=xthendydx=
Answered by mrW1 last updated on 06/Mar/17
xy=tan^(−1)  x  y=((tan^(−1)  x)/x)  (dy/dx)=−((tan^(−1)  x)/x^2 )+(1/(x(1+x^2 )))    way 2:  cos^2  (xy)=(1/(1+tan^2  (xy)))=(1/(1+x^2 ))    (1/(cos^2  (xy)))(y+x(dy/dx))=1  (dy/dx)=((cos^2  (xy)−y)/x)=(((1/(1+x^2 ))−y)/x)=(1/(x(1+x^2 )))−(y/x)=(1/(x(1+x^2 )))−((tan^(−1)  x)/x^2 )
xy=tan1xy=tan1xxdydx=tan1xx2+1x(1+x2)way2:cos2(xy)=11+tan2(xy)=11+x21cos2(xy)(y+xdydx)=1dydx=cos2(xy)yx=11+x2yx=1x(1+x2)yx=1x(1+x2)tan1xx2
Answered by bahmanfeshki last updated on 06/Mar/17
(xy′+y)(1+tan^2  (xy))=1  →y′=(1/x)((1/(1+x^2 ))−y)
(xy+y)(1+tan2(xy))=1y=1x(11+x2y)

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