Answer:
[tex]\frac{dy}{dx}=-\frac{2xe^{3-x^2}}{e^{3-x^2}+9}[/tex]
Explanation:
We were given the function:
[tex]y=\ln\left(e^{3-x^2}+9\right)[/tex]
We are to find the derivative of this function, we have it shown below:
[tex]\begin{gathered} y=\operatorname{\ln}(e^{3-x^{2}}+9) \\ If:y=ln(u) \\ Then:y^{\prime}=\frac{u^{\prime}}{u} \\ y^{\prime}=\frac{dy}{dy} \\ u=e^{3-x^2}+9 \\ u^{\prime}=-2xe^{3-x^2} \\ y^{\prime}=\frac{u^{\prime}}{u} \\ \Rightarrow y^{\prime}=\frac{-2xe^{3-x^2}}{e^{3-x^2}+9} \\ But:y^{\prime}=\frac{dy}{dx} \\ \frac{dy}{dx}=-\frac{2xe^{3-x^2}}{e^{3-x^2}+9} \\ \\ \therefore\frac{dy}{dx}=-\frac{2xe^{3-x^2}}{e^{3-x^2}+9} \end{gathered}[/tex]
