Question

use implicit differentiation to find \\(\frac{dy}{dx}\\)
\\x^3 + y^3 = 18xy\\
\\(\frac{dy}{dx} = \\)

Explanation:

Differentiate both sides with respect to x

$$ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(18xy) $$

$$ 3x^2 + 3y^2 \frac{dy}{dx} = 18y + 18x \frac{dy}{dx} $$

Isolate the derivative terms

$$ 3y^2 \frac{dy}{dx} - 18x \frac{dy}{dx} = 18y - 3x^2 $$

$$ (3y^2 - 18x) \frac{dy}{dx} = 18y - 3x^2 $$

Solve for dy/dx and simplify

$$ \frac{dy}{dx} = \frac{18y - 3x^2}{3y^2 - 18x} $$

$$ \frac{dy}{dx} = \frac{3(6y - x^2)}{3(y^2 - 6x)} = \frac{6y - x^2}{y^2 - 6x} $$

Answer:

Use implicit differentiation to find \(\frac{dy}{dx}\)

$$x^3+y^3=18xy$$

\(\frac{dy}{dx} =\) <blank>\(\frac{6y - x^2}{y^2 - 6x}\)</blank>