Differentiate $$ y=\arccos \left(e^{3 x}\right) $$ Answer: $$ \frac{d y}{d x}=\square $$

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To differentiate $$ y = \arccos \left(e^{3x}\right) $$, we'll use the chain rule combined with the derivative of the arccos function.

1. **Derivative of $$\arccos(u)$$:**
   $$
   \frac{d}{dx} \arccos(u) = -\frac{u'}{\sqrt{1 - u^2}}
   $$
   
2. **Identify $$ u $$ and its derivative:**
   $$
   u = e^{3x} \quad \text{so} \quad u' = 3e^{3x}
   $$
   
3. **Apply the chain rule:**
   $$
   \frac{dy}{dx} = -\frac{3e^{3x}}{\sqrt{1 - \left(e^{3x}\right)^2}} = -\frac{3e^{3x}}{\sqrt{1 - e^{6x}}}
   $$

**Final Answer:**
$$
\frac{dy}{dx} = -\frac{3\, e^{3x}}{\sqrt{1 - e^{6x}}}
$$
$$ \frac{dy}{dx} = -\frac{3\, e^{3x}}{\sqrt{1 - e^{6x}}} $$

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