If `r=[2varphi+cos^(2)(2varphi+pi//4)]^(1//2)`, then what is the value of the derivative of `dr//dvarphi` at `varphi=pi//4` ?

Options

A. `2(1^(1//2))/(pi+1)`
B. `2(1^(2))/(pi+1)`
C. `(2^(1//2))/(pi+1)`
D. `2(2^(1//2))/(pi+1)`

Answer

To find the value of the derivative \( \frac{dr}{d\varphi} \) at \( \varphi = \frac{\pi}{4} \) for the function

\[ r = \sqrt{2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)}, \]

we will follow these steps:

### Step 1: Differentiate \( r \) with respect to \( \varphi \)

Using the chain rule, we differentiate \( r \):

\[ \frac{dr}{d\varphi} = \frac{1}{2} \left(2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)\right)^{-1/2} \cdot \frac{d}{d\varphi}\left(2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)\right). \]

### Step 2: Differentiate the inner function

Now, we need to differentiate the inner function:

\[ \frac{d}{d\varphi}\left(2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)\right). \]

Using the chain rule again, we have:

\[ \frac{d}{d\varphi}\left(\cos^2\left(2\varphi + \frac{\pi}{4}\right)\right) = 2\cos\left(2\varphi + \frac{\pi}{4}\right)(-\sin\left(2\varphi + \frac{\pi}{4}\right) \cdot 2). \]

Thus,

\[ \frac{d}{d\varphi}\left(2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)\right) = 2 - 2\sin\left(2\varphi + \frac{\pi}{4}\right)\cos\left(2\varphi + \frac{\pi}{4}\right). \]

### Step 3: Substitute back into the derivative

Now substituting back, we have:

\[ \frac{dr}{d\varphi} = \frac{1}{2} \left(2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right)\right)^{-1/2} \cdot \left(2 - 2\sin\left(2\varphi + \frac{\pi}{4}\right)\cos\left(2\varphi + \frac{\pi}{4}\right)\right). \]

### Step 4: Evaluate at \( \varphi = \frac{\pi}{4} \)

Now we will evaluate \( \frac{dr}{d\varphi} \) at \( \varphi = \frac{\pi}{4} \):

1. Calculate \( 2\varphi + \cos^2\left(2\varphi + \frac{\pi}{4}\right) \):

\[ 2\left(\frac{\pi}{4}\right) + \cos^2\left(2\left(\frac{\pi}{4}\right) + \frac{\pi}{4}\right) = \frac{\pi}{2} + \cos^2\left(\frac{3\pi}{4}\right) = \frac{\pi}{2} + \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{2} + \frac{1}{2}. \]

2. Substitute this into the derivative:

\[ \frac{dr}{d\varphi} = \frac{1}{2} \left(\frac{\pi}{2} + \frac{1}{2}\right)^{-1/2} \cdot \left(2 - 2\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac{3\pi}{4}\right)\right). \]

3. Calculate \( \sin\left(\frac{3\pi}{4}\right) \) and \( \cos\left(\frac{3\pi}{4}\right) \):

\[ \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}. \]

4. Substitute these values:

\[ 2 - 2\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) = 2 + 1 = 3. \]

5. Finally, we have:

\[ \frac{dr}{d\varphi} = \frac{1}{2} \left(\frac{\pi}{2} + \frac{1}{2}\right)^{-1/2} \cdot 3. \]

### Final Step: Simplify

Thus, the value of \( \frac{dr}{d\varphi} \) at \( \varphi = \frac{\pi}{4} \) is:

\[ \frac{3}{2\sqrt{\frac{\pi}{2} + \frac{1}{2}}}. \]