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What is the slope of the tangent line of #e^(xy)-e^(x^2-x)/y = C #, where C is an arbitrary constant, at #(3,1)#?

Answer 1

Charles Anctil

For` #f(x,y) = C# we have the implicit function theorem:

#dy/dx = - f_x/f_y#

So:

#f_x = y e^(xy) - 1/ y(2x - 1) e^(x^2 - x)#

And:

#f_y = x e^(xy) + 1/y^2 e^(x^2 - x)#

So:

#dy/dx = - (y e^(xy) - 1/ y(2x - 1) e^(x^2 - x))/(x e^(xy) + 1/y^2 e^(x^2 - x))#

#= - (y^3 e^(xy) - y(2x - 1) e^(x^2 - x))/(xy^2 e^(xy) + e^(x^2 - x))#

#implies (dy/dx)_{ (3,1)} = - (e^(3) - 5 )/(3 e^(3) + 1)#

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Answer 2

Luke Alvarez

The slope of the tangent line of ( e^{xy} - \frac{e^{x^2 - x}}{y} = C ) at the point (3,1) can be found by taking the derivative of the function with respect to x, then evaluating it at the given point. This process involves implicit differentiation and the chain rule. After finding the derivative, substitute the given point's coordinates into the expression to find the slope of the tangent line.

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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.