If f(x)=x^{tan x}+(tan x)^{x}, then f^{prime} | vedclass

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(A) Let $f(x) = f_1(x) + f_2(x)$, where $f_1(x) = x^{	an x}$ and $f_2(x) = (	an x)^x$. For $f_1(x) = x^{	an x}$, taking log on both sides: $\log f_

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$1+\frac{\pi}{2} \log \left(\frac{e \pi}{4}\right)$

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(A) Let $f(x) = f_1(x) + f_2(x)$, where $f_1(x) = x^{	an x}$ and $f_2(x) = (	an x)^x$. For $f_1(x) = x^{	an x}$, taking log on both sides: $\log f_1 = 	an x \cdot \log x$. Differentiating with respect to $x$: $\frac{1}{f_1} \frac{df_1}{dx} = \sec^2 x \cdot \log x + 	an x \cdot \frac{1}{x}$. So, $\frac{df_1}{dx} = x^{	an x} \left( \sec^2 x \cdot \log x + \frac{	an x}{x} ight)$. For $f_2(x) = (	an x)^x$, taking log on both sides: $\log f_2 = x \cdot \log(	an x)$. Differentiating with respect to $x$: $\frac{1}{f_2} \frac{df_2}{dx} = 1 \cdot \log(	an x) + x \cdot \frac{1}{	an x} \cdot \sec^2 x$. So, $\frac{df_2}{dx} = (	an x)^x \left( \log(	an x) + \frac{x \sec^2 x}{	an x} ight)$. Now, $f^{\prime}(x) = \frac{df_1}{dx} + \frac{df_2}{dx}$. At $x = \frac{\pi}{4}$: $f_1^{\prime}\left(\frac{\pi}{4}ight) = \left(\frac{\pi}{4}ight)^1 \left( (\sqrt{2})^2 \log \frac{\pi}{4} + \frac{1}{\pi/4} ight) = \frac{\pi}{4} (2 \log \frac{\pi}{4} + \frac{4}{\pi}) = \frac{\pi}{2} \log \frac{\pi}{4} + 1$. $f_2^{\prime}\left(\frac{\pi}{4}ight) = (1)^{\pi/4} \left( \log 1 + \frac{\pi/4 \cdot 2}{1} ight) = 1 \cdot (0 + \frac{\pi}{2}) = \frac{\pi}{2}$. Therefore, $f^{\prime}\left(\frac{\pi}{4}ight) = \frac{\pi}{2} \log \frac{\pi}{4} + 1 + \frac{\pi}{2} = 1 + \frac{\pi}{2} (\log \frac{\pi}{4} + 1) = 1 + \frac{\pi}{2} \log \left( \frac{e \pi}{4} ight)$.

If $f(x)=x^{\tan x}+(\tan x)^{x}$, then $f^{\prime}\left(\frac{\pi}{4}\right)=$

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