How do you implicitly differentiate #-1=(x-y)sinx+y#?
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To implicitly differentiate the equation (-1 = (x - y) \sin(x) + y), follow these steps:
- Differentiate both sides of the equation with respect to (x).
- Apply the product rule and chain rule where necessary.
- Solve for (\frac{dy}{dx}) in terms of (x) and (y).
The steps are as follows:
- Differentiate both sides with respect to (x):
[ \frac{d}{dx}(-1) = \frac{d}{dx}[(x - y) \sin(x) + y] ]
- For the right side, use the sum rule and the product rule:
[ 0 = (x - y) \frac{d}{dx}[\sin(x)] + \sin(x) \frac{d}{dx}(x - y) + \frac{dy}{dx} ]
- Differentiate ( \sin(x) ) and (x - y) with respect to (x):
[ 0 = (x - y) \cos(x) + \sin(x) - \frac{dy}{dx} ]
- Solve for ( \frac{dy}{dx} ):
[ \frac{dy}{dx} = \sin(x) + (x - y) \cos(x) ]
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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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