use implicit differentiation to find $\frac{dy}{dx}$. $4y^{2}=\frac{3x - 2}{3x + 2}$ $\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $4y^{2}$ with respect to $x$ using chain - rule and $\frac{3x - 2}{3x+2}$ using quotient - rule. The derivative of $4y^{2}$ with respect to $x$ is $8y\frac{dy}{dx}$ (by chain - rule: $\frac{d}{dx}(4y^{2})=4\times2y\times\frac{dy}{dx}$). The quotient - rule states that if $u = 3x - 2$ and $v = 3x + 2$, then $\frac{d}{dx}(\frac{u}{v})=\frac{u'v - uv'}{v^{2}}$. Here, $u'=3$ and $v' = 3$. So, $\frac{d}{dx}(\frac{3x - 2}{3x + 2})=\frac{3(3x + 2)-3(3x - 2)}{(3x + 2)^{2}}$.

Step2: Simplify the right - hand side

Simplify $\frac{3(3x + 2)-3(3x - 2)}{(3x + 2)^{2}}$. [ \begin{align*} \frac{3(3x + 2)-3(3x - 2)}{(3x + 2)^{2}}&=\frac{9x+6-(9x - 6)}{(3x + 2)^{2}}\ &=\frac{9x+6 - 9x + 6}{(3x + 2)^{2}}\ &=\frac{12}{(3x + 2)^{2}} \end{align*} ] So, we have the equation $8y\frac{dy}{dx}=\frac{12}{(3x + 2)^{2}}$.

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

Isolate $\frac{dy}{dx}$ by dividing both sides of the equation $8y\frac{dy}{dx}=\frac{12}{(3x + 2)^{2}}$ by $8y$. $\frac{dy}{dx}=\frac{12}{8y(3x + 2)^{2}}=\frac{3}{2y(3x + 2)^{2}}$

Answer:

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