# 009A Sample Final 1, Problem 7

A curve is defined implicitly by the equation

$x^{3}+y^{3}=6xy.$ (a) Using implicit differentiation, compute  ${\frac {dy}{dx}}$ .

(b) Find an equation of the tangent line to the curve  $x^{3}+y^{3}=6xy$ at the point  $(3,3)$ .

Foundations:
1. What is the result of implicit differentiation of  $xy?$ It would be  $y+x{\frac {dy}{dx}}$ by the Product Rule.

2. What two pieces of information do you need to write the equation of a line?

You need the slope of the line and a point on the line.

3. What is the slope of the tangent line of a curve?

The slope is  $m={\frac {dy}{dx}}.$ Solution:

(a)

Step 1:
Using implicit differentiation on the equation  $x^{3}+y^{3}=6xy,$ we get

$3x^{2}+3y^{2}{\frac {dy}{dx}}=6y+6x{\frac {dy}{dx}}.$ Step 2:
Now, we move all the  ${\frac {dy}{dx}}$ terms to one side of the equation.
So, we have

$3x^{2}-6y={\frac {dy}{dx}}(6x-3y^{2}).$ We solve to get
${\frac {dy}{dx}}={\frac {3x^{2}-6y}{6x-3y^{2}}}.$ (b)

Step 1:
First, we find the slope of the tangent line at the point  $(3,3).$ We plug  $(3,3)$ into the formula for  ${\frac {dy}{dx}}$ we found in part (a).
So, we get

${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {3(3)^{2}-6(3)}{6(3)-3(3)^{2}}}\\&&\\&=&\displaystyle {-{\frac {9}{9}}}\\&&\\&=&\displaystyle {-1.}\end{array}}$ Step 2:
Now, we have the slope of the tangent line at  $(3,3)$ and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at  $(3,3)$ is

$y\,=\,-1(x-3)+3.$ (a)    ${\frac {dy}{dx}}={\frac {3x^{2}-6y}{6x-3y^{2}}}$ (b)    $y=-1(x-3)+3$ 