How do you differentiate #-y=lnxy-xy#?

Answer 1

#y' = - (y - xy^2)/( xy + x - x^2y )#

this is a bit easier [administratively] if you use the Implicit Function Theorem...

which says that #color{red}{y' = - (f_x)/(f_y)}# for function #f(x,y) = const#
so from #-y=lnxy-xy#
we get # y + lnxy-xy \color{blue}{= 0}= f(x,y)#
and so.... #f_x = 1/(xy)*y - y # #= 1/x - y# and #f_y = 1 + 1/(xy)*x - x # #= 1 + 1/(y) - x #
so using the IFT.... #y' = - (f_x)/(f_y)#
#= - (1/x - y)/( 1 + 1/(y) - x )#
#= - (y - xy^2)/( xy + x - x^2y )#

there are some pretty simple ways to intuit the Implicit Function Theorem

one is to note that if #f(x,y) = c#
then the total differential #df = 0#
and #df = f_x dx + f_y dy = 0#
so #dy/dx = - (f_x)/(f_y)#
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Answer 2

To differentiate the equation -y = ln(xy) - xy with respect to both x and y, you can use implicit differentiation.

Differentiate both sides of the equation with respect to x: d/dx(-y) = d/dx(ln(xy) - xy)

This yields: -dy/dx = (1/xy)(y + x(dy/dx)) - (y + x(dy/dx))

Now, differentiate both sides of the equation with respect to y: d/dy(-y) = d/dy(ln(xy) - xy)

This results in: -1 = (1/x)(y + x(dy/dx)) + x

Now, solve for dy/dx by isolating it in the equation obtained from differentiating with respect to x. This gives: dy/dx = -xy/(x + 1 + ln(xy))

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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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