Question

$\begin{aligned} &f(x)=x+\sin x \\ &g^{\prime}(x)=\frac{1}{1+\cos x} \\ &\text { Put } x=\pi / 4 \\ &g^{\prime}\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}}{1+\sqrt{2}}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2} \end{aligned}$

Short answer

Answer: \(2 - \sqrt{2}\)

Step-by-step solution

Simplify \(g'(x)\)

we are given \(g'(x) = \frac{1}{1 + \cos x}\). This expression is already in its simplest form, so no further simplification is required.

Substitute \(x = \frac{\pi}{4}\)

Now we need to substitute \(x = \frac{\pi}{4}\) into our expression for \(g'(x)\). This yields: \(g'\left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right) = \frac{1}{1 + \cos\left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right)}\)

Evaluate the expression

Using the addition formula for cosine, we can rewrite the expression as follows: \[\cos\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)=\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{1}{\sqrt{2}}\right)-\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}\sin\left(\frac{1}{\sqrt{2}}\right)\] Now, substituting this back into our expression for \(g'(x)\), we get: \[g'\left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right) = \frac{1}{1 + \frac{1}{\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}}\sin\left(\frac{1}{\sqrt{2}}\right)}\] We are given that: \[g'\left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right) = \frac{\sqrt{2}}{1 + \sqrt{2}} = \sqrt{2}(\sqrt{2} - 1) = 2 - \sqrt{2}\] Thus, we have found the value of \(g'\left( \frac{\pi}{4} + \frac{1}{\sqrt{2}} \right)\) to be \(2 - \sqrt{2}\).