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BrendanMcKee · Answer 1 · 2023-10-12T10:31:49+0000

Answer:

Step-by-step explanation:

We were given the function:

$y=\ln\left(e^(3-x^2)+9\right)$

We are to find the derivative of this function, we have it shown below:

$\begin{gathered} y=\operatorname{\ln}(e^{3-x^(2)}+9) \\ If:y=ln(u) \\ Then:y^(\prime)=(u^(\prime))/(u) \\ y^(\prime)=(dy)/(dy) \\ u=e^(3-x^2)+9 \\ u^(\prime)=-2xe^(3-x^2) \\ y^(\prime)=(u^(\prime))/(u) \\ \Rightarrow y^(\prime)=(-2xe^(3-x^2))/(e^(3-x^2)+9) \\ But:y^(\prime)=(dy)/(dx) \\ (dy)/(dx)=-(2xe^(3-x^2))/(e^(3-x^2)+9) \\ \\ \therefore(dy)/(dx)=-(2xe^(3-x^2))/(e^(3-x^2)+9) \end{gathered}$

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