# How do you implicitly differentiate #tan(x+y) = x#?

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To implicitly differentiate ( \tan(x+y) = x ), follow these steps:

- Use the chain rule to differentiate ( \tan(x+y) ) with respect to ( x ).
- Differentiate ( x ) with respect to ( x ).
- Solve for the derivative of ( y ) with respect to ( x ).

The result of implicitly differentiating ( \tan(x+y) = x ) with respect to ( x ) yields:

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

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

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

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