009A Sample Final A, Problem 4

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

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

Finding the slope:
We use implicit differentiation on our original equation to find
$-3x^{2}-2y-2x\cdot y'+3y^{2}\cdot y'=0.$
From here, I would immediately plug in (1,1) to find y ':
$-3-2-2y'+3y'=0$, or $y'=5.$

Writing the Equation of the Tangent Line:
Now, we simply plug our values of $x=y=1$ and $m=5$  into the point-slope form to find the tangent line through $(1,1)$ is
$y-1=5(x-1),$
or in slope-intercept form
$y=5x-4.$

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