Question 1027271

I'm going to use y to simplify,
{{{y^4=x^3+3x^2y+3xy^2+y^3}}}
Implicitly differentiating,
{{{4y^3dy=3x^2dx+3(x^2dy+2xydx)+3(y^2dx+2xydy)+3y^2dy}}}
{{{4y^3dy-3y^2dy-3x^2dy-6xydy)=3x^2dx+6xydx+3y^2dx}}}
{{{(4y^3-3y^2-6xy-3y^2)dy=3(x^2+2xy+y^2)dx}}}
{{{(4y^3-3y^2-6xy-3y^2)dy=3(x+y)^2dx}}}
{{{dy/dx=(3(x+y)^2)/(4y^3-3y^2-6xy-3y^2)}}}
However, it's not the case that {{{f(1)=2}}} because,
{{{(2)^4=(1+2)^3}}}
{{{16=(3)^3}}}
{{{16<>27}}}
Actually,
{{{y^4=(1+y)^3}}}
has two real solutions,
{{{y=-0.55}}} and {{{y=2.63}}}
which yields two values for the derivatives,
{{{dy/dx=-0.480}}} and {{{dy/dx=1.189}}}
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Here is the function along with the two tangent lines at {{{x=1}}}
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*[illustration gf2.JPG].