# 009A Sample Final A, Problem 4

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4. Find an equation for the tangent line to the function  ${\displaystyle -x^{3}-2xy+y^{3}=-1}$ at the point ${\displaystyle (1,1)}$.

Foundations:
Since only two variables are present, we are going to differentiate everything with respect to ${\displaystyle x}$ in order to find an expression for the slope, ${\displaystyle m=y'=dy/dx}$. Then we can use the point-slope equation form ${\displaystyle y-y_{1}=m(x-x_{1})}$ at the point ${\displaystyle \left(x_{1},y_{1}\right)=(1,1)}$ to find the equation of the tangent line.
Note that implicit differentiation will require the product rule and the chain rule. In particular, differentiating ${\displaystyle 2xy}$ can be treated as
${\displaystyle (2x)\cdot (y),}$
which has as a derivative
${\displaystyle 2\cdot y+2x\cdot y'=2y+2x\cdot y'.}$

Solution:

Finding the slope:
We use implicit differentiation on our original equation to find
${\displaystyle -3x^{2}-2y-2x\cdot y'+3y^{2}\cdot y'=0.}$
From here, I would immediately plug in ${\displaystyle (1,1)}$ to find ${\displaystyle y'}$:
${\displaystyle -3-2-2y'+3y'=0}$, or ${\displaystyle y'=5.}$
Writing the Equation of the Tangent Line:
Now, we simply plug our values of ${\displaystyle x=y=1}$ and ${\displaystyle m=5}$  into the point-slope form to find the tangent line through ${\displaystyle (1,1)}$ is
${\displaystyle y-1=5(x-1),}$
or in slope-intercept form
${\displaystyle y=5x-4.}$