Express 208'x - 3 sin(x) in the form R cos(x + α).

Question

ScanSolve AI Tutor · Accepted Answer

Let's differentiate the given expression. From the image, the expression appears to be $x \cos(2x) + \frac{1}{2} \sin(2x)$ .

Identify the function to differentiate.
Let $y = x \cos(2x) + \frac{1}{2} \sin(2x)$ .
We need to find $\frac{dy}{dx}$ .
Differentiate the first term, $x \cos(2x)$ , using the product rule.
The product rule states that $\frac{d}{dx}(uv) = u'v + uv'$ .
Let $u = x$ and $v = \cos(2x)$ .
Then $u' = \frac{d}{dx}(x) = 1$ .
To find $v'$ , we use the chain rule: $\frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -\sin(2x) \cdot 2 = -2\sin(2x)$ .
Applying the product rule:
$\frac{d}{dx}(x \cos(2x)) = (1)(\cos(2x)) + (x)(-2\sin(2x))$ $\frac{d}{dx}(x \cos(2x)) = \cos(2x) - 2x\sin(2x)$
Differentiate the second term, $\frac{1}{2} \sin(2x)$ , using the chain rule.
$\frac{d}{dx}\left(\frac{1}{2} \sin(2x)\right) = \frac{1}{2} \cdot \frac{d}{dx}(\sin(2x))$
Using the chain rule: $\frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos(2x)$ .
So,
$\frac{d}{dx}\left(\frac{1}{2} \sin(2x)\right) = \frac{1}{2} \cdot (2\cos(2x))$ $\frac{d}{dx}\left(\frac{1}{2} \sin(2x)\right) = \cos(2x)$
Combine the results from Step 2 and Step 3.
$\frac{dy}{dx} = (\cos(2x) - 2x\sin(2x)) + (\cos(2x))$ $\frac{dy}{dx} = \cos(2x) + \cos(2x) - 2x\sin(2x)$ $\frac{dy}{dx} = 2\cos(2x) - 2x\sin(2x)$
The final answer is $2\cos(2x) - 2x\sin(2x)$ .

Answer