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:

$\begin{array}{rl}{x}^4-x^{2}y+y^4& =1\\ (x^4-x^{2}y+y^4{)}^{\prime }& =(1{)}^{\prime }\\ 4x^3-(x^{2}y{)}^{\prime }+4y^3{y}^{\prime }& =0\\ 4x^3-x^2{y}^{\prime }-2xy+4y^3{y}^{\prime }& =0\\ y^{\prime }(-x^2+4y^3)& =2xy-4x^3\\ y^{\prime }& =\frac{2xy-4x^3}{4y^3-x^2}\end{array}$

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:

$\begin{array}{rl}{y}^{\prime }& =\frac{2xy-4x^3}{4y^3-x^2}\\ & =\frac{2(-1)(1)-4(-1{)}^3}{4(1{)}^3-(-1{)}^2}\\ & =\frac{-2+4}{4-1}\\ & =\frac{2}{3}\\ \end{array}$

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