QuestionLet $y=\ln \left(x^{2}+y^{2}\right)$. Determine the derivative $y^{\prime}$ at the point $\left(\sqrt{e^{3}-9}, 3\right)$. $$y^{\prime}\left(\sqrt{e^{3}-9}\right)=$$

Studdy Solution
Simplify the expression:
Calculate $(\sqrt{e^3 - 9})^2$:
$$(\sqrt{e^3 - 9})^2 = e^3 - 9$$
Substitute back:
$$y' = \frac{2\sqrt{e^3 - 9}}{e^3 - 9 + 9 - 6}$$
$$y' = \frac{2\sqrt{e^3 - 9}}{e^3 - 6}$$
The derivative $y'$ at the point $\left(\sqrt{e^3 - 9}, 3\right)$ is:
$$\boxed{\frac{2\sqrt{e^3 - 9}}{e^3 - 6}}$$