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

Question

Isabella Ackerman · Accepted Answer

#y'=(y/x-1-ln(xy))/((x)/y-1-ln(xy))#

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

#18=(y-x)ln (xy)#

#d/dx(18)=d/dx((y-x)*ln (yx))#

#0=(y-x)*d/dx(ln (xy))+ln (xy)*d/dx(y-x)#

#0=(y-x)*(1/(xy))*d/dx(xy)+(ln (xy))(y'-1)#

#0=(y-x)*(xy'+y*1)/(xy)+y'ln(xy)-ln(xy)#

#0=(y-x)*((y')/y+1/x)+y'ln(xy)-ln(xy)#

#0=y'+y/x-(xy')/y-1+y'ln(xy)-ln(xy)#

Transpose the terms with y'

#(xy')/y-y'-y'ln(xy)=y/x-1-ln(xy)#

factor out the y'

#((x)/y-1-ln(xy))*y'=y/x-1-ln(xy)#

Divide both sides by #((x)/y-1-ln(xy))#

#(cancel((x)/y-1-ln(xy))*y')/cancel(((x)/y-1-ln(xy)))=(y/x-1-ln(xy))/((x)/y-1-ln(xy))#

#y'=(y/x-1-ln(xy))/((x)/y-1-ln(xy))#

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

Cameron Alexander · Answer

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

1. Differentiate each term with respect to $x$.
2. Use the product rule for differentiating $\ln(xy)$.
3. 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:

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

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

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