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

Answer 1

Ezekiel Alexander

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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Answer 2

Charles Arocha

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

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

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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.