Questions: If (f(x)=4 cos ^2(x)), compute its differential (d f). (d f=) Approximate the change in (f) when (x) changes from (x=fracpi6) to (x=fracpi6+0.1). (Round your answer to three decimal places.) [ Delta f approx ]

Solution Steps

To solve the given problem, we need to find the differential of the function \( f(x) = 4 \cos^2(x) \) and then approximate the change in \( f \) when \( x \) changes from \( \frac{\pi}{6} \) to \( \frac{\pi}{6} + 0.1 \).

1. Find the differential \( df \):

   • Use the chain rule to differentiate \( f(x) = 4 \cos^2(x) \).
   • The derivative of \( \cos^2(x) \) is \( 2 \cos(x) \cdot (-\sin(x)) \) using the chain rule.
   • Therefore, \( f'(x) = 4 \cdot 2 \cos(x) \cdot (-\sin(x)) = -8 \cos(x) \sin(x) \).
   • The differential \( df \) is then \( df = f'(x) \cdot dx = -8 \cos(x) \sin(x) \cdot dx \).

2. Approximate the change in \( f \):

   • Use the differential to approximate the change in \( f \) when \( x \) changes from \( \frac{\pi}{6} \) to \( \frac{\pi}{6} + 0.1 \).
   • Calculate \( df \) at \( x = \frac{\pi}{6} \) and \( dx = 0.1 \).

To find the differential of the function \( f(x) = 4 \cos^2(x) \), we first compute its derivative. Using the chain rule, we have:

\[ f'(x) = -8 \cos(x) \sin(x) \]

Thus, the differential \( df \) is given by:

\[ df = f'(x) \cdot dx = -8 \cos(x) \sin(x) \cdot dx \]

Next, we evaluate \( df \) at \( x = \frac{\pi}{6} \) with a change \( dx = 0.1 \):

1. Calculate \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \).
2. Calculate \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

Substituting these values into the derivative:

\[ f'\left(\frac{\pi}{6}\right) = -8 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = -2\sqrt{3} \]

Now, substituting into the differential:

\[ df = -2\sqrt{3} \cdot 0.1 = -0.2\sqrt{3} \]

To approximate the change in \( f \) when \( x \) changes from \( \frac{\pi}{6} \) to \( \frac{\pi}{6} + 0.1 \), we compute:

\[ \Delta f \approx df = -0.2\sqrt{3} \approx -0.3464 \]

Rounding this to three decimal places gives:

\[ \Delta f \approx -0.346 \]

Final Answer