How do you find #(dy)/(dx)# given #4xy+4x+3=0#?

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Answer 1 · 2016-11-03T01:18:05

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

I propose a slightly different approach.

#4xy = -3 - 4x#

#xy = (-3 - 4x)/4#

Now, differentiate with respect to #x#.

#y + x(dy/dx) = (-4 xx 4 - 0 xx (-3 - 4x))/4^2#

#y + x(dy/dx) = -16/16#

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

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

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

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

Hopefully this helps!