Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (g)/Solution 1

First, we must find the derivative of the curve using implicit differentiation:

$\displaystyle {\begin{aligned}x^{4}-x^{2}y+y^{4}&=1\\(x^{4}-x^{2}y+y^{4})'&=(1)'\\4x^{3}-(x^{2}y)'+4y^{3}y'&=0\\4x^{3}-x^{2}y'-2xy+4y^{3}y'&=0\\y'(-x^{2}+4y^{3})&=2xy-4x^{3}\\y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\end{aligned}}$

This gives us the equation for the slope. The question is asking us for the slope at the point (-1,1) so we plug in x=-1 and y=1:

$\displaystyle {\begin{aligned}y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\\&={\frac {2(-1)(1)-4(-1)^{3}}{4(1)^{3}-(-1)^{2}}}\\&={\frac {-2+4}{4-1}}\\&={\frac {2}{3}}\\\end{aligned}}$

Therefore, the slope of the tangent line to the curve at (-1,1) is 2/3.