# 022 Sample Final A, Problem 4

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

Foundations:
When we use implicit differentiation, we combine the chain rule with the fact that ${\displaystyle y}$ 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
${\displaystyle {\frac {d}{dx}}\left(y^{3}\right)=3y^{2}\cdot {\frac {dy}{dx}}.}$
For this problem, we also need to use the product rule.

Solution:

Step 1:
First, we differentiate each term separately with respect to ${\displaystyle x}$ and apply the product rule on the right hand side to find that  ${\displaystyle x+y=x^{3}y^{3}}$  differentiates implicitly to
${\displaystyle 1+{\frac {dy}{dx}}\,=\,3x^{2}y^{3}+3x^{3}y^{2}\cdot {\frac {dy}{dx}}}$.
Step 2:
Now we need to solve for  ${\displaystyle {\frac {dy}{dx}}}$ , and doing so we find that  ${\displaystyle {\frac {dy}{dx}}\,=\,{\frac {3x^{2}y^{3}-1}{1-3x^{3}y^{2}}}}$.
Final Answer:
${\displaystyle {\frac {dy}{dx}}\,=\,{\frac {3x^{2}y^{3}-1}{1-3x^{3}y^{2}}}.}$