# How do you find the slope of a tangent line to the graph of the function #x^3 + y^3 – 6xy = 0#, at (4/3, 8/3)?

use implicit differentiation

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To find the slope of a tangent line to the graph of a function at a specific point, we can use the derivative of the function.

First, we need to find the derivative of the function. Taking the derivative of x^3 + y^3 – 6xy = 0 with respect to x, we get 3x^2 + 3y^2(dy/dx) - 6y - 6x(dy/dx) = 0.

Next, we can substitute the given point (4/3, 8/3) into the equation and solve for dy/dx. Plugging in the values, we have 3(4/3)^2 + 3(8/3)^2(dy/dx) - 6(8/3) - 6(4/3)(dy/dx) = 0.

Simplifying the equation, we get 4 + 64(dy/dx) - 16 - 32(dy/dx) = 0.

Combining like terms, we have 32(dy/dx) + 48 = 0.

Solving for dy/dx, we get dy/dx = -48/32 = -3/2.

Therefore, the slope of the tangent line to the graph of the function x^3 + y^3 – 6xy = 0 at the point (4/3, 8/3) is -3/2.

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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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