How do you implicitly differentiate #18=(y-x)^2-ln(xy)#?

Answer 1

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

#18=(y-x)^2-ln(xy)#
#0=-2(y-x)+2(y-x)dy/dx - (y/(xy)+x/(xy) dy/dx)#
#0=-2(y-x)+2(y-x)dy/dx - 1/x-1/y dy/dx#
#2(y-x)+1/x=2(y-x)dy/dx-1/y dy/dx#
#2(y-x)+1/x=(2(y-x)-1/y )dy/dx#
#(2(y-x)+1/x)/(2(y-x)-1/y )=dy/dx#
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Answer 2

To implicitly differentiate (18=(y-x)^2-\ln(xy)), follow these steps:

  1. Differentiate each term of the equation with respect to (x).
  2. Apply the chain rule when differentiating composite functions.
  3. Solve for (\frac{dy}{dx}) by isolating the terms involving (y) on one side and the terms involving (x) on the other side.
  4. Express (\frac{dy}{dx}) in terms of (x) and (y).

Differentiating each term: [ \begin{align*} \frac{d}{dx}(18) &= 0 \ \frac{d}{dx}((y-x)^2) &= 2(y-x)\frac{dy}{dx} - 2 \ \frac{d}{dx}(\ln(xy)) &= \frac{1}{xy}(y + x\frac{dy}{dx}) \end{align*} ]

Substituting into the equation and solving for (\frac{dy}{dx}): [ 0 = 2(y-x)\frac{dy}{dx} - 2 - \frac{1}{xy}(y + x\frac{dy}{dx}) ]

[ 2(y-x)\frac{dy}{dx} - \frac{y}{xy} - \frac{x}{xy}\frac{dy}{dx} = 2 ]

[ (2(y-x) - \frac{x}{y})\frac{dy}{dx} = 2 + \frac{y}{xy} ]

[ \frac{dy}{dx} = \frac{2 + \frac{y}{xy}}{2(y-x) - \frac{x}{y}} ]

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Answer 3

To implicitly differentiate (18 = (y - x)^2 - \ln(xy)) with respect to (x), you'll differentiate each term with respect to (x) while treating (y) as a function of (x) using the chain rule where necessary. Then, solve for (\frac{dy}{dx}).

Starting with the given equation:

[ 18 = (y - x)^2 - \ln(xy) ]

Differentiating each term:

  1. Differentiate ((y - x)^2): [ \frac{d}{dx}(y - x)^2 = 2(y - x) \frac{dy}{dx} - 1 ]

  2. Differentiate (-\ln(xy)): [ \frac{d}{dx}(-\ln(xy)) = -\frac{1}{xy} \cdot (y + xy') - \frac{1}{xy} \cdot (x \cdot y' + y) ]

Combining terms:

[ 0 = 2(y - x) \frac{dy}{dx} - 1 - \frac{y}{x} - y' - \frac{x}{y} - x \cdot y' - y ]

Simplify and solve for (y'):

[ 0 = 2(y - x) \frac{dy}{dx} - 1 - \frac{y}{x} - y' - \frac{x}{y} - x \cdot y' - y ]

[ 0 = 2(y - x) \frac{dy}{dx} - y' - x \cdot y' ]

[ 0 = (2(y - x) - 1 - x) \frac{dy}{dx} - y' ]

[ \frac{dy}{dx} = \frac{y'}{2(y - x) - 1 - x} ]

That's the implicit differentiation of the given equation with respect to (x), giving (\frac{dy}{dx}) in terms of (x), (y), and (y').

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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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