Answer to Question #328984 in Calculus for Val

Question

Find dy/dx by implicit differentiation

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

Answer 1

$\tan(x-y)=\frac{y}{1+x^2}$

$(\tan(x-y))’=(\frac{y}{1+x^2})’$

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

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

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

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