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