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Find the derivative of the function. \[ g(x)=9 \arccos \left(\frac{x}{6}\right) \] \( g^{\prime}(x)=\square \)

Ask by Weber Tyler. in the United States
Feb 26,2025

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Answer

The derivative of \( g(x) = 9 \arccos\left(\frac{x}{6}\right) \) is \( g'(x) = -\frac{3}{2\sqrt{1 - \left(\frac{x}{6}\right)^2}} \) or equivalently \( g'(x) = -\frac{9}{\sqrt{36 - x^2}} \).

Solution

We are given the function   g(x) = 9 · arccos(x/6). Step 1. Recall that the derivative of arccos(u) with respect to x is   d/dx [arccos(u)] = –u' / √(1 – u²). Step 2. Here, u = x/6 so that u' = 1/6. Step 3. Differentiate using the chain rule:   g'(x) = 9 · d/dx [arccos(x/6)]      = 9 · [ –(1/6) / √(1 – (x/6)²) ]      = –9/(6√(1 – (x/6)²)). Step 4. Simplify the constant:   –9/(6√(1 – (x/6)²)) = –3/(2√(1 – (x/6)²)). Thus, the derivative is:   g'(x) = –3/(2√(1 – (x/6)²)). Alternatively, since √(1 – (x/6)²) = √((36 – x²)/36) = √(36 – x²)/6, you can also write:   g'(x) = –3/(2 · (√(36 – x²)/6)) = –9/√(36 – x²). Both forms are equivalent.

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The Deep Dive

To find the derivative of the function \( g(x) = 9 \arccos\left(\frac{x}{6}\right) \), we can apply the chain rule. The derivative of \( \arccos(u) \) is \( -\frac{1}{\sqrt{1-u^2}} \), where \( u = \frac{x}{6} \). First, we find the derivative of \( \frac{x}{6} \), which is \( \frac{1}{6} \). Now, applying the chain rule, we have: \[ g'(x) = 9 \cdot \left(-\frac{1}{\sqrt{1-\left(\frac{x}{6}\right)^2}} \right) \cdot \frac{1}{6} \] Simplifying this gives us: \[ g'(x) = -\frac{9}{6\sqrt{1 - \frac{x^2}{36}}} = -\frac{3}{2\sqrt{1 - \frac{x^2}{36}}} \] So, the final answer is: \[ g'(x) = -\frac{3}{2\sqrt{1 - \frac{x^2}{36}}} \]

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