Find $$ y^{\prime} $$ if $$ y=\ln \left(9 x^{2}+5 y^{2}\right) $$ $$ y^{\prime}=\square $$

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Find the first order derivative with respect to $$ x $$ for $$ y=\ln(9x^2+5y^2) $$.
Find the first derivative by following steps:
- step0: Find the derivative with respect to $$x$$:
  $$y=\ln{\left(9x^{2}+5y^{2}\right)}$$
- step1: Take the derivative:
  $$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\ln{\left(9x^{2}+5y^{2}\right)}\right)$$
- step2: Calculate the derivative:
  $$\frac{dy}{dx}=\frac{d}{dx}\left(\ln{\left(9x^{2}+5y^{2}\right)}\right)$$
- step3: Calculate the derivative:
  $$\frac{dy}{dx}=\frac{18x+10y\frac{dy}{dx}}{9x^{2}+5y^{2}}$$
- step4: Cross multiply:
  $$\frac{dy}{dx}\times \left(9x^{2}+5y^{2}\right)=18x+10y\frac{dy}{dx}$$
- step5: Simplify the equation:
  $$\left(9x^{2}+5y^{2}\right)\frac{dy}{dx}=18x+10y\frac{dy}{dx}$$
- step6: Move the variable to the left side:
  $$\left(9x^{2}+5y^{2}\right)\frac{dy}{dx}-10y\frac{dy}{dx}=18x$$
- step7: Collect like terms:
  $$\left(9x^{2}+5y^{2}-10y\right)\frac{dy}{dx}=18x$$
- step8: Divide both sides:
  $$\frac{\left(9x^{2}+5y^{2}-10y\right)\frac{dy}{dx}}{9x^{2}+5y^{2}-10y}=\frac{18x}{9x^{2}+5y^{2}-10y}$$
- step9: Divide the numbers:
  $$\frac{dy}{dx}=\frac{18x}{9x^{2}+5y^{2}-10y}$$
The derivative of $$ y $$ with respect to $$ x $$ is given by:

$$ y' = \frac{18x}{9x^{2}+5y^{2}-10y} $$

Therefore, the derivative of $$ y $$ with respect to $$ x $$ is $$ \frac{18x}{9x^{2}+5y^{2}-10y} $$.
$$ y' = \frac{18x}{9x^{2}+5y^{2}-10y} $$

Find \( y^{\prime} \) if \( y=\ln \left(9 x^{2}+5 y^{2}\right) \) \( y^{\prime}=\square \)

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