Question #328984

Find dy/dx by implicit differentiation

tan(x-y)=y/1+x^2


1
Expert's answer
2022-04-16T04:13:04-0400

tan(xy)=y1+x2\tan(x-y)=\frac{y}{1+x^2}

(tan(xy))=(y1+x2)(\tan(x-y))’=(\frac{y}{1+x^2})’

(xy)cos2(xy)=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}

ycos2(xy)=y(1+x2)2xy(1+x2)2\frac{-y’}{\cos^2(x-y)}=\frac{y’(1+x^2)-2xy}{(1+x^2)^2}

y(1+x2)2=(y(1+x2)2xy)cos2(xy)-y’(1+x^2)^2=(y’(1+x^2)-2xy) \cos^2(x-y)

y=2xycos2(xy)(1+x2)(cos2(xy)+(1+x2))y’=\frac{2xy\cos^2(x-y)}{(1+x^2)(\cos^2(x-y)+(1+x^2))}


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