How do you implicitly differentiate #-1=xy+cot^2(x-y) #?
so we get
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To implicitly differentiate the equation ( -1 = xy + \cot^2(x-y) ), follow these steps:
- Differentiate both sides of the equation with respect to ( x ).
- Use the product rule for the term ( xy ) and the chain rule for the term ( \cot^2(x-y) ).
- 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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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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