use implicit differentiation to find $\frac{dy}{dx}$. $x^{3}=\frac{4x + 3y}{4x-3y}$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}=\frac{4x + 3y}{4x-3y}$ with respect to $x$. The derivative of $x^{3}$ is $3x^{2}$. For the right - hand side, use the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u = 4x + 3y$ and $v=4x - 3y$. So $u'=4 + 3\frac{dy}{dx}$ and $v'=4-3\frac{dy}{dx}$. Then $\left(\frac{4x + 3y}{4x-3y}\right)'=\frac{(4 + 3\frac{dy}{dx})(4x-3y)-(4x + 3y)(4 - 3\frac{dy}{dx})}{(4x-3y)^{2}}$.

Step2: Simplify the right - hand side

Expand the numerator: [ \begin{align*} &(4 + 3\frac{dy}{dx})(4x-3y)-(4x + 3y)(4 - 3\frac{dy}{dx})\ =&16x-12y+12x\frac{dy}{dx}-9y\frac{dy}{dx}-(16x-12x\frac{dy}{dx}+12y - 9y\frac{dy}{dx})\ =&16x-12y+12x\frac{dy}{dx}-9y\frac{dy}{dx}-16x + 12x\frac{dy}{dx}-12y + 9y\frac{dy}{dx}\ =&24x\frac{dy}{dx}-24y \end{align*} ] So we have $3x^{2}=\frac{24x\frac{dy}{dx}-24y}{(4x-3y)^{2}}$.

Step3: Solve for $\frac{dy}{dx}$

Cross - multiply: $3x^{2}(4x - 3y)^{2}=24x\frac{dy}{dx}-24y$. Rearrange to get $24x\frac{dy}{dx}=3x^{2}(4x - 3y)^{2}+24y$. Then $\frac{dy}{dx}=\frac{3x^{2}(4x - 3y)^{2}+24y}{24x}$.

Answer:

$\frac{3x^{2}(4x - 3y)^{2}+24y}{24x}$