# What is the derivative of # tan(x-y)=y/(1+x^2)#?

Now take note of this:

so:

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To find the derivative of the given equation ( \tan(x-y) = \frac{y}{1+x^2} ), differentiate both sides with respect to ( x ) using the chain rule and product rule:

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

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

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

Solve for ( \frac{dy}{dx} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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