What is the slope of the tangent line of #e^(xy)/(2x-y)= C #, where C is an arbitrary constant, at #(2,3)#?

Answer 1

I would use the fact that #C# is not arbitrary, but is determined by the fact that #(2,3)# lies on the curve.

Since #(2,3)# lies on the curve, we can find #C = e^6#.

The equation becomes:

#e^(xy)= 2e^6x-e^6y# # # # # (for #y != 2x#)

Differentiating implicitly yields:

#e^(xy)(y+x dy/dx)=2e^6-e^6 dy/dx#
We were not asked for an explicit formula for #dy/dx#, so let's not bother with that.
At the point of interest #x=2# and #y=3#, so we get:
#e^6(3+2 dy/dx)= 2e^6-e^6 dy/dx#

So we need only solve:

#3+2 dy/dx = 2-dy/dx# for #dy/dx = -1/3# and we're done.
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Answer 2

To find the slope of the tangent line at the point (2,3) for the equation ( \frac{e^{xy}}{2x-y}= C ), we need to find the derivative of (y) with respect to (x), and then evaluate it at the point (2,3).

Differentiate both sides of the equation implicitly with respect to (x), and then solve for (y'), the derivative of (y) with respect to (x). Once you have (y'), substitute (x = 2) and (y = 3) into (y') to find the slope of the tangent line at the point (2,3).

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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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