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Find [tex]\frac{dy}{dx}[/tex] by Implicit Differentiation:

[tex]\tan(x - y) = \frac{y}{1 + x^2}[/tex]

[tex]\frac{d}{dx} (\tan(x - y)) = \frac{d}{dx} \left( \frac{y}{1 + x^2} \right)[/tex]

[tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - y \frac{d}{dx} (1 + x^2)}{(1 + x^2)^2} = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

[tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

What am I doing wrong?

Any suggestions?...

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# Implicit Differentation Help!

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