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 pointslope equation form $yy_{1}=m(xx_{1})$ 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

$2\cdot y+2x\cdot y'=2y+2x\cdot y'.$

Solution:
Finding the slope:

We use implicit differentiation on our original equation to find

$3x^{2}2y2x\cdot y'+3y^{2}\cdot y'=0.$

From here, I would immediately plug in $(1,1)$ to find $y'$:

$322y'+3y'=0$, 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 pointslope form to find the tangent line through $(1,1)$ is

$y1=5(x1),$

or in slopeintercept form

$y=5x4.$

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