use implicit differentiation to find dy/dx. 3y^2=(2x - 5)/(2x + 5) dy/dx=□

Explanation:

Step1: Differentiate both sides

Differentiate $3y^{2}$ and $\frac{2x - 5}{2x+5}$ with respect to $x$. The derivative of $3y^{2}$ using the chain - rule is $6y\frac{dy}{dx}$. For the right - hand side, use the quotient rule. If $u = 2x - 5$ and $v=2x + 5$, then $\frac{d}{dx}(\frac{u}{v})=\frac{u'v - uv'}{v^{2}}$, where $u'=2$ and $v' = 2$. So $\frac{d}{dx}(\frac{2x - 5}{2x+5})=\frac{2(2x + 5)-2(2x - 5)}{(2x + 5)^{2}}=\frac{4x+10-(4x - 10)}{(2x + 5)^{2}}=\frac{20}{(2x + 5)^{2}}$. So we have $6y\frac{dy}{dx}=\frac{20}{(2x + 5)^{2}}$.

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

Divide both sides of the equation $6y\frac{dy}{dx}=\frac{20}{(2x + 5)^{2}}$ by $6y$ (assuming $y\neq0$). $\frac{dy}{dx}=\frac{20}{6y(2x + 5)^{2}}=\frac{10}{3y(2x + 5)^{2}}$.

Answer:

$\frac{10}{3y(2x + 5)^{2}}$