find $\frac{dy}{dx}$ by implicit differentiation.\n$x^{3}+y^{3}=4$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}+y^{3}=4$ with respect to $x$. The derivative of $x^{3}$ with respect to $x$ is $3x^{2}$ using the power - rule. For $y^{3}$, by the chain - rule, its derivative with respect to $x$ is $3y^{2}\frac{dy}{dx}$. The derivative of the constant 4 with respect to $x$ is 0. So we have: $3x^{2}+3y^{2}\frac{dy}{dx}=0$

Step2: Solve for $\frac{dy}{dx}$

First, subtract $3x^{2}$ from both sides of the equation: $3y^{2}\frac{dy}{dx}=- 3x^{2}$ Then divide both sides by $3y^{2}$ (assuming $y\neq0$): $\frac{dy}{dx}=-\frac{x^{2}}{y^{2}}$

Answer:

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