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

Start from the given equation and differentiate both sides of the equation with respect to x

Transpose the terms with y'

factor out the y'

God bless....I hope the explanation is useful.

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

- Differentiate each term with respect to (x).
- Use the product rule for differentiating (\ln(xy)).
- Solve for (\frac{{dy}}{{dx}}) after differentiation.

Differentiating each term:

(0 = (y - x) \cdot \frac{{d}}{{dx}}(\ln(xy)) + \ln(xy) \cdot \frac{{d}}{{dx}}(y - x))

Applying the product rule:

(0 = (y - x) \cdot \left(\frac{{1}}{{xy}} \cdot (xy)' + \ln(xy) \cdot (y)' + \ln(xy) \cdot (x)'\right) + \ln(xy) \cdot (y)' - \ln(xy) \cdot (x)')

Simplify:

(0 = \frac{{y + xy'}}{{xy}} + \ln(xy) \cdot \frac{{dy}}{{dx}} + \ln(xy) - \frac{{x + xy'}}{{xy}})

Rearrange terms:

(\frac{{xy + x \cdot \frac{{dy}}{{dx}}}}{{xy}} = \frac{{y - \ln(xy)}}{{\ln(xy)}})

(\frac{{1 + \frac{{dy}}{{dx}}}}{{1}} = \frac{{y - \ln(xy)}}{{xy}})

(\frac{{dy}}{{dx}} = \frac{{y - \ln(xy) - 1}}{{x + y}})

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