022 Sample Final A, Problem 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
${\frac {d}{dx}}\left(y^{3}\right)=3y^{2}\cdot {\frac {dy}{dx}}.$
For this problem we also need to use the product rule.

Solution:

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

Step 2:

Now we need to solve for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}} and doing so we find that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}}

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