How do you differentiate #e^sqrt(xy)-sqrt(xy)=8#?

Answer 1

See explanation.

#dy/dx = -c/(x^2)#

where #c# is the square of the positive root of #e^t-t-8 = 0#

Let #t = sqrt(xy)#
Then #e^t - t = 8#
This has two roots #t_1 ~~ 2.33559#, #t_2 ~~ -7.999664#
Numerical approximations for #t_1# and #t_2# can be found using Newton's method or similar.
We can discard #t_2# since #t = sqrt(xy)# is the non-negative square root.
Let #c = t_1^2#

Then the original equation simplifies to:

#xy = c#
So #y = c/x# and #(dy)/(dx) = -c/(x^2)#
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Answer 2

To differentiate ( e^{\sqrt{xy}} - \sqrt{xy} = 8 ):

  1. Differentiate both sides of the equation with respect to (x).
  2. Use the chain rule and the power rule for differentiation.
  3. Solve for the derivative ( \frac{dy}{dx} ).

The result after differentiation is:

( \frac{1}{2}e^{\sqrt{xy}} \left(\frac{y}{2\sqrt{xy}} + \frac{x}{2\sqrt{xy}} \right) - \frac{1}{2\sqrt{xy}} - \frac{1}{2}\frac{x}{\sqrt{xy}} = 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.

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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