Question 377490
{{{x^3-y^2+xy-x^2=0}}}
{{{1^3-1^2+(1)(1)-1^2=0}}}
{{{1-1+1-1=0}}}
{{{0+0=0}}}
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True, so {{{1}}}{{{1}}}) is on the curve.
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Use implicit differentiation,
{{{x^3-y^2+xy-x^2=0}}}
{{{3x^2-2y(dy/dx)+x(dy/dx)+y-2x=0}}}
{{{(-2y+x)(dy/dx)=-3x^2+2x-y}}}
{{{dy/dx=(-3x^2+2x-y)/(x-2y)}}}
When {{{x=y=1}}}
{{{dy/dx=(-3+2-1)/(1-2)}}}
{{{dy/dx=(-2)/(-1)}}}
{{{highlight(dy/dx=2)}}}
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For the second derivative, use the quotient rule,
{{{r=p/q}}}
{{{dr/dx=(q*(dp/dx)-p*(dq/dx))/q^2}}}
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{{{dy/dx=(-3x^2+2x-y)/(x-2y)}}}
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{{{d(dy/dx)/dx=((x-2y)(-6x+2-dy/dx)-(-3x^2+2x-y)(1-2*(dy/dx)))/(x-2y)^2}}}
The value when {{{x=y=1}}} and {{{dy/dx=2}}} is,
{{{d(dy/dx)/dx=((1-2)(-6+2-2)-(-3+2-1)(1-2*(2)))/(1-2)^2}}}
{{{d(dy/dx)/dx=((-1)(-6)-(-2)(-3))/(-1)^2}}}
{{{d(dy/dx)/dx=(6-6)/1}}}
{{{highlight(d(dy/dx)/dx=0)}}}