How do you differentiate #5x-y=xyln(xy)#?

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Answer 1 · 2017-03-05T22:39:51

#dy/dx= (5- y- yln(xy))/(1+x+ xln(xy) )#

Differentiate both sides of the equation with respect to #x#:

#d/dx ( 5x -y) = d/dx (xyln(xy))#

Use the product rule at the second member:

#5-y' = d/dx(xy) ln(xy) + (xy) d/dx (ln(xy))#

Use the product rule again for #d/dx(xy) # and the chain rule for #d/dx (ln(xy))#:

#5-y' = (y+xy') ln(xy) + (xy) 1/(xy) d/dx (xy)#

Simplify:

#5-y' = (y+xy') ln(xy) + d/dx (xy)#

#5-y' = (y+xy') ln(xy) +(y+xy')#

#5-y' = (y+xy')(1+ ln(xy))#

Solve now for #y'#:

#5-y' = y(1+ ln(xy))+xy'(1+ ln(xy))#

#5- y(1+ ln(xy))=xy'(1+ ln(xy)) +y'#

#5- y(1+ ln(xy))=y'(x+ xln(xy) +1)#

#y' = (5- y- yln(xy))/(x+ xln(xy) +1)#