# How do you find the slope of a tangent line to the graph of the function #3xy-2x+3y^2=5# at (2,1)?

Please the explanation for steps leading to the slope,

I will differentiate each term, separately.

1.1 Use the product rule for term 1:

1.2 Use the power rule on term 2:

1.3 Use the chain rule on term 3:

1.4 Term 4 is a constant, 5, the derivative is zero.

Put these back into the equation:

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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. First, we need to find the derivative of the given function with respect to x. Then, we can substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. In this case, the derivative of the function 3xy - 2x + 3y^2 = 5 with respect to x is 3y - 2 + 6xy. Substituting the x-coordinate 2 into the derivative, we get the slope of the tangent line at (2,1) as 3y - 2 + 6xy = 3(1) - 2 + 6(2)(1) = 3 - 2 + 12 = 13. Therefore, the slope of the tangent line to the graph of the function 3xy - 2x + 3y^2 = 5 at (2,1) is 13.

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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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