Question

question number 6. (4.00 points)
given $f(x) = 6\sin(4x) + 2\cos(3x)$, find $f(x)$.
$\circ f(x) = 24 \cos(4 x) - 6 \sin(3 x)$

$\circ f(x) = 6 \cos(4 x) - 2 \sin(3 x)$

$\circ f(x) = -24 \cos(4 x) + 6 \sin(3 x)$

$\circ f(x) = 24 \cos(4 x) + 6 \sin(3 x)$

$\circ f(x) = 6 \cos(4 x) + 2 \sin(3 x)$

$\circ$none of the above.

Explanation:

Step1: Differentiate \(6\sin(4x)\)

Use the chain rule: \(\frac{d}{dx}[\sin(u)] = \cos(u) \cdot u'\). Here \(u = 4x\), so \(u' = 4\).
\(\frac{d}{dx}[6\sin(4x)] = 6 \cdot \cos(4x) \cdot 4 = 24\cos(4x)\)

Step2: Differentiate \(2\cos(3x)\)

Use the chain rule: \(\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot u'\). Here \(u = 3x\), so \(u' = 3\).
\(\frac{d}{dx}[2\cos(3x)] = 2 \cdot (-\sin(3x)) \cdot 3 = -6\sin(3x)\)

Step3: Combine the derivatives

\(f'(x) = 24\cos(4x) - 6\sin(3x)\)

Answer:

\(f'(x) = 24 \cos(4 x) - 6 \sin(3 x)\) (the first option)