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The problem given is to find the derivative of the function:

\[
y = 27 \arctan(\sqrt{x})
\]

### Step 1: Identify the Function Components
The function involves a constant multiplier and a composition of functions. Let's rewrite it as:
\[
y = 27 \cdot \arctan(u), \quad \text{where } u = \sqrt{x}
\]

### Step 2: Differentiate with Respect to \(x\)
We'll use the chain rule for differentiation:
\[
\frac{dy}{dx} = 27 \cdot \frac{d}{du} \left( \arctan(u) \right) \cdot \frac{du}{dx}
\]

### Step 3: Differentiate the Inner Functions
1. The derivative of \(\arctan(u)\) with respect to \(u\) is:
   \[
   \frac{d}{du} \left( \arctan(u) \right) = \frac{1}{1 + u^2}
   \]
2. The derivative of \(u = \sqrt{x} = x^{1/2}\) with respect to \(x\) is:
   \[
   \frac{du}{dx} = \frac{1}{2\sqrt{x}}
   \]

### Step 4: Substitute and Simplify
Now, substitute these derivatives into the expression:
\[
\frac{dy}{dx} = 27 \cdot \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}}
\]

Simplify the expression:
\[
\frac{dy}{dx} = 27 \cdot \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}
\]
\[
\frac{dy}{dx} = \frac{27}{2\sqrt{x}(1 + x)}
\]

### Final Answer
The derivative of the function \(y = 27 \arctan(\sqrt{x})\) with respect to \(x\) is:
\[
\frac{dy}{dx} = \frac{27}{2\sqrt{x}(1 + x)}
\]

Would you like more details on the steps, or do you have any questions?

Here are some related questions you might find interesting:
1. How do you apply the chain rule when differentiating more complex functions?
2. What are some common derivatives involving inverse trigonometric functions like \(\arctan(x)\)?
3. How do you differentiate expressions with square roots, such as \(\sqrt{x}\)?
4. What are the real-world applications of derivatives involving inverse trigonometric functions?
5. How do the derivatives of other inverse trigonometric functions, like \(\arcsin(x)\) and \(\arccos(x)\), compare?

**Tip:** When differentiating composite functions, always clearly identify the inner and outer functions to apply the chain rule efficiently.

Find the derivative of the function: y = 27 arctan(√x).

Learn how to differentiate the function y = 27 arctan(√x) with a step-by-step solution. This problem involves the chain rule and differentiation of inverse trigonometric functions, suitable for undergraduate calculus or advanced high school calculus students.

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