# How do you find the slope of the tangent line to the curve #6x^2 - 2xy + 3y^3 = 216# at the point 6,0?

Use implicit differentiation with respect to

By the way, it is worth confirming that

Also, we are assuming

Here's a picture of the situation in this problem. The tangent line at the given point is shown as well (dashed).

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To find the slope of the tangent line to the curve (6x^2 - 2xy + 3y^3 = 216) at the point ( (6,0) ), follow these steps:

- Differentiate the given equation implicitly with respect to (x) to find ( \frac{dy}{dx} ).
- Evaluate ( \frac{dy}{dx} ) at the point ( (6,0) ) to find the slope of the tangent line.

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To find the slope of the tangent line to the curve 6x^2 - 2xy + 3y^3 = 216 at the point (6,0), first, differentiate the equation implicitly with respect to x. Then, substitute the x-coordinate (6) and y-coordinate (0) into the resulting expression. Solve for dy/dx to find the slope of the tangent line at the given point.

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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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