# How do you implicitly differentiate #-1=xy+cot^2(x/y) #?

see the pic

by using partial differnetiation we differentiate implicit function

in last multiply numerator with -1 i forget to multiply

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To implicitly differentiate the equation -1 = xy + cot^2(x/y), we differentiate each term with respect to x, using the chain rule where necessary.

The derivative of -1 with respect to x is 0.
The derivative of xy with respect to x is y + x(dy/dx).
The derivative of cot^2(x/y) with respect to x is -2*cot(x/y) csc^2(x/y)(1/y - x/y^2)*(dy/dx).

Setting up the equation and solving for dy/dx yields:
0 = y + x(dy/dx) - 2*cot(x/y) csc^2(x/y)(1/y - x/y^2)*(dy/dx)

dy/dx = (2*cot(x/y) csc^2(x/y)(1/y - x/y^2) - y) / (x - x*y/y^2)

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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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