How do you use implicit differentiation to find the slope of the curve given #x^2y=x+2# at (2,1)?
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Please see the explanation for the procedure. The slope is
I will explain each term.
Term 2:
Term 3:
Put these back into their corresponding locations in the equation:
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To find the slope of the curve at the point (2,1) using implicit differentiation:
 Differentiate both sides of the equation with respect to x.
 Solve for dy/dx.
 Plug in the values of x and y from the point (2,1) into the expression for dy/dx to find the slope at that point.
Starting with the given equation:
x^2y = x + 2

Differentiate both sides with respect to x:
d/dx (x^2y) = d/dx (x + 2)

Apply the product rule on the left side and the derivative of a constant on the right side:
2xy + x^2(dy/dx) = 1

Solve for dy/dx:
dy/dx = (1  2xy) / x^2

Plug in the values of x=2 and y=1:
dy/dx = (1  221) / 2^2 = (1  4) / 4 = 3/4
So, the slope of the curve at the point (2,1) is 3/4.
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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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