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

Answer 1

#y'=(2cot(x-y)-ysin(x-y))/(xsin(x-y)+2cot(x-y))#

#0=y+xy'-2cot(x-y)/sin(x.y)(1-y')#
#0=y-2cot(x-y)/sin(x-y)+y'(x+2cot(x-y)/sin(x-y))# Multiplying by
#sin(x-y)#
#2cot(x-y)-ysin(x-y)=y'(xsin(x-y)+2cot(x-y))#

so we get

#y'=(2cos(x-y)-ysin(x-y))/(xsin(x-y)+2cot(x-y))#
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Answer 2

To implicitly differentiate the equation ( -1 = xy + \cot^2(x-y) ), follow these steps:

  1. Differentiate both sides of the equation with respect to ( x ).
  2. Use the product rule for the term ( xy ) and the chain rule for the term ( \cot^2(x-y) ).
  3. After differentiation, solve for ( \frac{dy}{dx} ), which represents the derivative of ( y ) with respect to ( x ).

The result is ( \frac{dy}{dx} = \frac{y - \sin(2x)}{x + \sin(2x)} ).

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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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