# How do you find #dy/dx# by implicit differentiation of #xy=4# and evaluate at point (-4,-1)?

Let's differentiate both sides

graph{(y+2+x/4)(y-4/x)=0 [-17.28, 2.73, -4.94, 5.06]}

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To find ( \frac{dy}{dx} ) by implicit differentiation of ( xy = 4 ), follow these steps:

- Differentiate both sides of the equation with respect to ( x ).
- Use the product rule for differentiation on the left side.
- Solve the resulting equation for ( \frac{dy}{dx} ).
- Substitute the given point ( (-4, -1) ) into the expression for ( \frac{dy}{dx} ) to evaluate.

Starting with ( xy = 4 ):

- ( \frac{d}{dx}(xy) = \frac{d}{dx}(4) )
- Apply the product rule: ( y + x\frac{dy}{dx} = 0 )
- Solve for ( \frac{dy}{dx} ): ( \frac{dy}{dx} = -\frac{y}{x} )
- Substitute ( (-4, -1) ): ( \frac{dy}{dx} = -\frac{-1}{-4} = \frac{1}{4} )

So, ( \frac{dy}{dx} = \frac{1}{4} ) when evaluated at the point ( (-4, -1) ).

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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.

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