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What is the derivative of #tan(x − y) = x#?

Answer 1

Charles Anctil

#dy/dx=x^2/(1+x^2)#.

#tan(x-y)=x#

#rArr x-y=arctanx#

#rArr x-arctanx=y#

#rArr dy/dx=1-1/(1+x^2)=x^2/(1+x^2)#.

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Answer 2

Paisley Johnson

To find the derivative of the given equation ( \tan(x - y) = x ) with respect to ( x ), we will use implicit differentiation.

Differentiating both sides of the equation with respect to ( x ):

[ \frac{d}{dx}(\tan(x - y)) = \frac{d}{dx}(x) ]

Using the chain rule and the derivative of tangent function:

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

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

Now, solving for ( \frac{dy}{dx} ):

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

This is the derivative of ( \tan(x - y) ) with respect to ( x ).

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Answer from HIX Tutor

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.