How do you differentiate #xy=cot(xy)#?

Answer 1

# dy/dx = - y/x #

I am assuming that you want to find # dy/dx #.
We shall use both implicit differentiation and chain rule. # xy = cot(xy) #
Differentiate both sides with respect to #x# to get: # d/dx (xy) = d/dx (cot(xy)) #
Apply chain rule # d/dx (xy) = -csc^2(xy) d/dx (xy) # # => (1 + csc^2(xy)) d/dx (xy) = 0 # # => d/dx (xy) = 0 # (Since #(1 + csc^2(xy))# is always positive)
Apply chain rule again # y + x dy/dx = 0 # # dy/dx = - y/x #
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Answer 2

To differentiate ( xy = \cot(xy) ), you would use implicit differentiation. Differentiate both sides of the equation with respect to ( x ) and apply the product rule on the left side and the chain rule on the right side.

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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.

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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