How do you find #dy/dx# by implicit differentiation given #x=siny+cosy#?
# dy/dx = 1/(cos y -sin y) #
We have:
Method 1 - Use the Chain Rule
Method 2 - Implicit Differentiation
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To find ( \frac{dy}{dx} ) by implicit differentiation given ( x = \sin(y) + \cos(y) ), follow these steps:
- Differentiate both sides of the equation with respect to ( x ).
- Use the chain rule when differentiating terms involving ( y ).
- Solve for ( \frac{dy}{dx} ).
Differentiating ( x = \sin(y) + \cos(y) ) with respect to ( x ) yields:
[ \frac{dx}{dx} = \frac{d}{dx}(\sin(y) + \cos(y)) ]
[ 1 = \frac{d}{dy}(\sin(y)) \frac{dy}{dx} + \frac{d}{dy}(\cos(y)) \frac{dy}{dx} ]
[ 1 = \cos(y) \frac{dy}{dx} - \sin(y) \frac{dy}{dx} ]
[ \frac{dy}{dx} (\cos(y) - \sin(y)) = 1 ]
[ \frac{dy}{dx} = \frac{1}{\cos(y) - \sin(y)} ]
This is the derivative ( \frac{dy}{dx} ) in terms of ( y ).
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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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