022 Exam 1 Sample A, Problem 2

2. Use implicit differentiation to find $dy/dx$ at the point $(1,0)$ on the curve defined by $x^{3}-y^{3}-y=x$.

ExpandFoundations:
When we use implicit differentiation, we combine the chain rule with the fact that $y$ is a function of $x$, and could really be written as $y(x).$ Because of this, the derivative with respect to $x$ of $y^{3}$ requires the chain rule, so
${\frac {d}{dx}}\left(y^{3}\right)=3y^{2}\cdot {\frac {dy}{dt}}.$

 Solution:

ExpandStep 1:
First, we differentiate each term separately with respect to x to find that  $x^{3}-y^{3}-y=x$  differentiates implicitly to
$3x^{2}-3y^{2}\cdot {\frac {dy}{dx}}-{\frac {dy}{dx}}=1$.

ExpandStep 2:
Since they don't ask for a general expression of $dy/dx$, but rather a particular value at a particular point, we can plug in the values $x=1$ and $y=0$  to find
$3(1)^{2}-3(0)^{2}\cdot {\frac {dy}{dx}}-{\frac {dy}{dx}}=1,$
which is equivalent to $3-{\frac {dy}{dx}}=1$. This solves to $dy/dx=2.$

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