How do you find #dy/dx# by implicit differentiation given #xy^2-3x^2y+x=1#?

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Answer 1 · 2017-09-17T06:54:06

The answer is #=((6xy-y^2-1))/((2yx-3x^2))#

We need

#(uv)'=u'v+uv'#

The function is

#xy^2-3x^2y+x-1=0#

We differentiate with respect to x

#d/(dx)(xy^2)-d/(dx)(3x^2y)+d/(dx)(x)-d/(dx)(1)=0#

#y^2+2yxdy/dx-6xy-3x^2dy/dx+1=0#

#dy/dx(2yx-3x^2)=6xy-y^2-1#

#dy/dx=((6xy-y^2-1))/((2yx-3x^2))#