022 Sample Final A, Problem 4

4. Use implicit differentiation to find ${\frac {dy}{dx}}:\qquad x+y=x^{3}y^{3}$

Foundations:
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 of $y^{3}$ with respect to $x$ requires the chain rule, so
${\frac {d}{dx}}\left(y^{3}\right)=3y^{2}\cdot {\frac {dy}{dt}}.$

 Solution:

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

Step 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 $dy/dx=2.$

Final Answer:
$dy/dx=2.$

Return to Sample Exam