How do you find the slope of the tangent line to the curve #4xy^3+3xy=7# at the point (1,1) by using imlicit differentiation?

Answer 1

I found: Slope#=-7/15#

First you need to derive implicitly (rememberin that #y# will be a function of #x#) and get (using the Product Rule on the two terms on the left): #4y^3+12xy^2(dy)/(dx)+3y+3x(dy)/(dx)=0# collect #(dy)/(dx)#: #(dy)/(dx)=(-4y^3-3y)/(12xy^2+3x)=#Slope Now use the coordinates of your point: #x=1# #y=1# #(dy)/(dx)=(-4-3)/(12+3)=-7/15#
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Answer 2

To find the slope of the tangent line using implicit differentiation, differentiate the equation with respect to x, then solve for dy/dx. The equation is (4xy^3 + 3xy = 7). Differentiating with respect to x yields (4y^3 + 12xy^2(dy/dx) + 3y + 3x(dy/dx) = 0). Plug in the point (1,1) to find the value of dy/dx, which represents the slope of the tangent line at that point.

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Answer from HIX Tutor

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.

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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