# 009A Sample Final 3, Problem 5

Calculate the equation of the tangent line to the curve defined by  ${\displaystyle x^{3}+y^{3}=2xy}$  at the point,  ${\displaystyle (1,1).}$

Foundations:
The equation of the tangent line to  ${\displaystyle f(x)}$  at the point  ${\displaystyle (a,b)}$  is
${\displaystyle y=m(x-a)+b}$  where  ${\displaystyle m=f'(a).}$

Solution:

Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
${\displaystyle 3x^{2}+3y^{2}y'=2y+2xy'.}$
We rearrange the terms and solve for  ${\displaystyle y'.}$
Therefore,
${\displaystyle 3x^{2}-2y=2xy'-3y^{2}y'}$
and
${\displaystyle y'={\frac {3x^{2}-2y}{2x-3y^{2}}}.}$
Step 2:
Therefore, the slope of the tangent line at the point  ${\displaystyle (1,1)}$  is
${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {3(1)^{2}-2(1)}{2(1)-3(1)^{2}}}\\&&\\&=&\displaystyle {\frac {3-2}{2-3}}\\&&\\&=&\displaystyle {-1.}\end{array}}}$
Hence, the equation of the tangent line to the curve at the point  ${\displaystyle (1,1)}$  is
${\displaystyle y=-1(x-1)+1.}$

${\displaystyle y=-1(x-1)+1}$