How do you implicitly differentiate #11=(x+y)/(xe^y+ye^x)#?

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Answer 1 · 2018-07-27T21:30:57

#dy/dx=\frac{11ye^x+11e^y-1}{1-11xe^y-11e^x}#

Given function:

#11={x+y}/{xe^y+ye^x}#

#11xe^y+11ye^x=x+y#

Differentiating above equation w.r.t. #x# on both the sides as follows

#d/dx(11xe^y+11ye^x)=d/dx(x+y)#

#11xe^ydy/dx+11e^y+11ye^x+11e^xdy/dx=1+dy/dx#

#(1-11xe^y-11e^x)dy/dx=11ye^x+11e^y-1#

#dy/dx=\frac{11ye^x+11e^y-1}{1-11xe^y-11e^x}#