Questions: Let f(t) = sqrt(6) / t^5. Then, f'(t) = □ f'(2) = □

Transcript text: Let $f(t)=\frac{\sqrt{6}}{t^{5}}$. Then, \[ \begin{array}{l} f^{\prime}(t)=\square \\ f^{\prime}(2)=\square \end{array} \]

Solution

Solution Steps

To solve this problem, we need to find the derivative of the function $ f(t) = \frac{\sqrt{6}}{t^5} $ and then evaluate this derivative at $ t = 2 $.

Find the derivative $ f'(t) $:
- Use the power rule for differentiation. Rewrite the function as $ f(t) = \sqrt{6} \cdot t^{-5} $.
- Differentiate using the power rule: $ \frac{d}{dt} [t^n] = n \cdot t^{n-1} $.
Evaluate the derivative at $ t = 2 $:
- Substitute $ t = 2 $ into the derivative to find $ f'(2) $.

Step 1: Define the Function

Given the function: \[ f(t) = \frac{\sqrt{6}}{t^5} \]

Step 2: Differentiate the Function

To find the derivative $ f'(t) $, we rewrite the function as: \[ f(t) = \sqrt{6} \cdot t^{-5} \] Using the power rule $ \frac{d}{dt} [t^n] = n \cdot t^{n-1} $, we get: \[ f'(t) = \sqrt{6} \cdot (-5) \cdot t^{-6} = -5 \cdot \frac{\sqrt{6}}{t^6} \]

Step 3: Evaluate the Derivative at $ t = 2 $

Substitute $ t = 2 $ into the derivative: \[ f'(2) = -5 \cdot \frac{\sqrt{6}}{2^6} = -5 \cdot \frac{\sqrt{6}}{64} \]

Final Answer

The derivative of the function is: \[ f'(t) = -5 \cdot \frac{\sqrt{6}}{t^6} \] Evaluating this at $ t = 2 $: \[ f'(2) = -5 \cdot \frac{\sqrt{6}}{64} \]

\[ \boxed{f'(t) = -5 \cdot \frac{\sqrt{6}}{t^6}} \] \[ \boxed{f'(2) = -5 \cdot \frac{\sqrt{6}}{64}} \]

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