How do you use implicit differentiation to find dy/dx given #xy+y=sinx#?
Now differentiate using the product rule on the first term.
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To use implicit differentiation to find ( \frac{dy}{dx} ) given ( xy + y = \sin(x) ), follow these steps:
 Differentiate both sides of the equation with respect to ( x ).
 Treat ( y ) as a function of ( x ) and use the product rule for differentiating ( xy ).
 Apply the chain rule to differentiate ( \sin(x) ).
 Solve the resulting equation for ( \frac{dy}{dx} ).
Let's go through the steps:

Differentiating both sides: [ \frac{d}{dx}(xy + y) = \frac{d}{dx}(\sin(x)) ]

Applying the product rule to ( xy ): [ \frac{d(xy)}{dx} + \frac{dy}{dx} = \cos(x) ]

Differentiating ( \sin(x) ): [ \frac{d}{dx}(\sin(x)) = \cos(x) ]

Substitute the results back into the original equation and solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} + x\frac{dy}{dx} + y = \cos(x) ] [ \frac{dy}{dx}(1 + x) = \cos(x)  y ] [ \frac{dy}{dx} = \frac{\cos(x)  y}{1 + x} ]
So, ( \frac{dy}{dx} = \frac{\cos(x)  y}{1 + x} ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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