Question

What is the derivative of the function
$f\left(x\right)=e^{\tan x}+\ln \left(\sec x\right)-e^{\ln x}atx=\frac{\pi }{4}$

• A. $e/2$
• B. $e$
• C. $2e$
• D. $4e$

Answer 1

The correct option is C $2e$

Given $f\left(x\right)=e^{tanx}+ln\left(secx\right)-e^{lnx}$

$\Rightarrow f\left(x\right)=e^{tanx}+ln\left(secx\right)-x$

Differentiating with respect to $x$ we get

$f^{\prime }\left(x\right)=e^{tanx}\cdot se{c}^{2}x+\frac{1}{secx}\cdot \left(secxtanx\right)-1$

Now we have to find $f^{\prime }\left(\frac{\pi }{4}\right)$

We know that $tan\frac{\pi }{4}=1,sec\frac{\pi }{4}=\sqrt{2}$

$\therefore f^{\prime }\left(\frac{\pi }{4}\right)=e^1\cdot (\sqrt{2}{)}^2+\frac{1}{\sqrt{2}}\cdot (\sqrt{2}\cdot 1)-1$

$=2e+1-1$

$=2e$

Hence $f^{\prime }\left(\frac{\pi }{4}\right)=2e$