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

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Answer 1 · 2016-07-01T11:43:23

#(xy-y)/(x-xy)#

Given expression

#ln(xy)=x+y #

#=>lnx+lny=x+y#

Differentiating w.r. to x we can write

#(d(lnx))/(dx)+(d(lny))/(dx)=(d(x))/(dx)+(d(y))/(dx)#

#=>1/x+1/y*(dy)/(dx)=1+(dy)/(dx)#

#=>1/y*(dy)/(dx)-(dy)/(dx)=1-1/x#

#=>(1/y-1)(dy)/(dx)=(x-1)/x#

#=>((1-y)/y)(dy)/(dx)=(x-1)/x#

#=>(dy)/(dx)=(x-1)/x xx(y)/(1-y)=(xy-y)/(x-xy)#