How do you differentiate #e^(xy) + x^2 - y^2 = 10#?

Answer 1

Elijah Abram

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

#e^(xy) + x^2 - y^2 = 10 = f(x,y)#

#f_x = y e^{xy} + 2x#

#f_y = x e^{xy} -2y#

by implicit function theorem: #dy/dx = - (f_x)/(f_y) = - (y e^{xy} + 2x)/(x e^{xy} -2y)#

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Answer 2

William Abdalla

To differentiate ( e^{xy} + x^2 - y^2 = 10 ) with respect to ( x ) and ( y ):

Differentiate with respect to ( x ): [ \frac{\partial}{\partial x} (e^{xy}) + \frac{\partial}{\partial x} (x^2) - \frac{\partial}{\partial x} (y^2) = 0 ] [ y e^{xy} + 2x = 0 ]

Differentiate with respect to ( y ): [ \frac{\partial}{\partial y} (e^{xy}) + \frac{\partial}{\partial y} (x^2) - \frac{\partial}{\partial y} (y^2) = 0 ] [ x e^{xy} - 2y = 0 ]

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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.