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Find dy/dx by implicit differentiation. tan(x-y)=y/(5+x^2)
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Differentiate both sides of the equation:\[\sec^2(x-y)(1-y')=\frac{y'(5+x^2)-2xy}{(5+x^2)^2}\]and solve for y': \[\sec^2(x-y)(1-y')(5+x^2)^2=y'(5+x^2)-2xy\]\[y'(5+x^2+\sec^2(x-y)(5+x^2)^2)=\sec^2(x-y)\left(5+x^2\right)^2+2xy\]\[y' = \frac{\sec^2(x-y)(5+x^2)^2+2xy}{\left(5+x^2\right)\left(1+\sec^2(x-y)(5+x^2)\right)}.\]
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