Find dy/dx by implicit differentiation
tan(x-y)=y/1+x^2
tan(x−y)=y1+x2\tan(x-y)=\frac{y}{1+x^2}tan(x−y)=1+x2y
(tan(x−y))’=(y1+x2)’(\tan(x-y))’=(\frac{y}{1+x^2})’(tan(x−y))’=(1+x2y)’
(x−y)’cos2(x−y)=y’(1+x2)−y(1+x2)’(1+x2)2\frac{(x-y)’}{\cos^2(x-y)}=\frac{y’(1+x^2)-y(1+x^2)’}{(1+x^2)^2}cos2(x−y)(x−y)’=(1+x2)2y’(1+x2)−y(1+x2)’
−y’cos2(x−y)=y’(1+x2)−2xy(1+x2)2\frac{-y’}{\cos^2(x-y)}=\frac{y’(1+x^2)-2xy}{(1+x^2)^2}cos2(x−y)−y’=(1+x2)2y’(1+x2)−2xy
−y’(1+x2)2=(y’(1+x2)−2xy)cos2(x−y)-y’(1+x^2)^2=(y’(1+x^2)-2xy) \cos^2(x-y)−y’(1+x2)2=(y’(1+x2)−2xy)cos2(x−y)
y’=2xycos2(x−y)(1+x2)(cos2(x−y)+(1+x2))y’=\frac{2xy\cos^2(x-y)}{(1+x^2)(\cos^2(x-y)+(1+x^2))}y’=(1+x2)(cos2(x−y)+(1+x2))2xycos2(x−y)
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