# How do you use differentials and the function #f(x,y) = arctan(x*y^2)# to approximate the value of f(0.94, 1.17)?

The tangent plane is given by:

And so we have:

Revisiting our equation for the tangent plane:

Putting in our values:

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To approximate the value of ( f(0.94, 1.17) ) using differentials, we'll use the total differential formula:

[ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy ]

First, we find the partial derivatives of ( f(x, y) = \arctan(xy^2) ) with respect to ( x ) and ( y ):

[ \frac{\partial f}{\partial x} = \frac{y^2}{1 + (xy^2)^2} ] [ \frac{\partial f}{\partial y} = \frac{2xy}{1 + (xy^2)^2} ]

Then, we'll evaluate these partial derivatives at the point ( (0.94, 1.17) ) and compute ( dx ) and ( dy ) by taking the differences between the given point and the nearby point ( (x, y) ):

[ dx = 0.94 - 0.94 = 0 ] [ dy = 1.17 - 1.17 = 0 ]

Now, we'll substitute the values into the formula to find the approximate value of ( df ):

[ df = \frac{1.17^2}{1 + (0.94 \times 1.17^2)^2} \times 0 + \frac{2 \times 0.94 \times 1.17}{1 + (0.94 \times 1.17^2)^2} \times 0 ]

[ df = 0 ]

Therefore, the approximate value of ( f(0.94, 1.17) ) using differentials is ( 0 ).

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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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