# 022 Exam 1 Sample A, Problem 2

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

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

Solution:

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