If r = [2phi + cos^2(2phi + pi/4)]^{1/2},then | vedclass

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(D) Given $r = [2\phi + \cos^2(2\phi + \pi/4)]^{1/2}$.Applying the chain rule,we differentiate with respect to $\phi$:$\frac{dr}{d\phi} = \frac{1}{2}[

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$2\left( \frac{2}{\pi + 1} \right)^{1/2}$

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(D) Given $r = [2\phi + \cos^2(2\phi + \pi/4)]^{1/2}$.Applying the chain rule,we differentiate with respect to $\phi$:$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot \frac{d}{d\phi}[2\phi + \cos^2(2\phi + \pi/4)]$$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot [2 + 2\cos(2\phi + \pi/4) \cdot (-\sin(2\phi + \pi/4)) \cdot 2]$$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot [2 - 2\sin(4\phi + \pi/2)]$At $\phi = \pi/4$,the argument $2\phi + \pi/4 = \pi/2 + \pi/4 = 3\pi/4$.$\cos^2(3\pi/4) = (-1/\sqrt{2})^2 = 1/2$.$\sin(4(\pi/4) + \pi/2) = \sin(\pi + \pi/2) = \sin(3\pi/2) = -1$.Substituting these values:$\frac{dr}{d\phi} = \frac{1}{2}[2(\pi/4) + 1/2]^{-1/2} \cdot [2 - 2(-1)]$$\frac{dr}{d\phi} = \frac{1}{2}[\pi/2 + 1/2]^{-1/2} \cdot [4]$$\frac{dr}{d\phi} = 2 \cdot [(\pi + 1)/2]^{-1/2} = 2 \cdot [2/(\pi + 1)]^{1/2}$.

If $r = [2\phi + \cos^2(2\phi + \pi/4)]^{1/2}$,then what is the value of the derivative $dr/d\phi$ at $\phi = \pi/4$?

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