Question

if (y = ln(2x^{2}-3y^{2})), then (\frac{dy}{dx}=)
a (\frac{1}{2x^{2}-3y^{2}})
b (\frac{4x}{2x^{2}-3y^{2}})
c (\frac{4x - 6y}{2x^{2}-3y^{2}})
d (\frac{4x}{2x^{2}-3y^{2}+6y})

Explanation:

Step1: Recall chain - rule for differentiation

If $y = \ln(u)$, then $\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}$. Here $u = 2x^{2}-3y^{2}$.

Step2: Differentiate both sides of $y=\ln(2x^{2}-3y^{2})$ with respect to $x$

Using the chain - rule, the left - hand side is $\frac{dy}{dx}$, and the right - hand side is $\frac{1}{2x^{2}-3y^{2}}\cdot\frac{d}{dx}(2x^{2}-3y^{2})$.

Step3: Differentiate $2x^{2}-3y^{2}$ with respect to $x$

$\frac{d}{dx}(2x^{2}-3y^{2})=\frac{d}{dx}(2x^{2})-\frac{d}{dx}(3y^{2})$. We know that $\frac{d}{dx}(2x^{2}) = 4x$ and by the chain - rule $\frac{d}{dx}(3y^{2})=6y\frac{dy}{dx}$.

Step4: Substitute the derivative of $2x^{2}-3y^{2}$ back into the equation

$\frac{dy}{dx}=\frac{4x - 6y\frac{dy}{dx}}{2x^{2}-3y^{2}}$.

Step5: Cross - multiply and solve for $\frac{dy}{dx}$

$(2x^{2}-3y^{2})\frac{dy}{dx}=4x - 6y\frac{dy}{dx}$.
$(2x^{2}-3y^{2})\frac{dy}{dx}+6y\frac{dy}{dx}=4x$.
$\frac{dy}{dx}(2x^{2}-3y^{2}+6y)=4x$.
$\frac{dy}{dx}=\frac{4x}{2x^{2}-3y^{2}+6y}$.

Answer:

D. $\frac{4x}{2x^{2}-3y^{2}+6y}$