# Thread: did i do this correctly?

1. ## did i do this correctly?

find $f'(\frac{\pi}{4})$
if $f(x)=\frac{2\sqrt{x}}{tan x}$

my answer: $f'(\frac{\pi}{4})=\frac{2-4\pi}{\sqrt{\pi}}$

2. Originally Posted by yoman360
find $f'(\frac{\pi}{4})$
if $f(x)=\frac{2\sqrt{x}}{tan x}$

my answer: $f'(\frac{\pi}{4})=\frac{2-4\pi}{\sqrt{\pi}}$
Close. The derivative is of $f$ is:

$\frac{\tan x\cdot\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot\sec^2x}{\tan^2x}$

Plugging in $\frac{\pi}{4}$ gives us

$\frac{\tan(\pi/4)\cdot\frac{1}{\sqrt{\pi/4}}-2\sqrt{\pi/4}\cdot\sec^2(\pi/4)}{\tan^2(\pi/4)}=\frac{\frac{2}{\sqrt{\pi}}-\sqrt{\pi}\cdot2}{1} = \boxed{\frac{2-2\pi}{\sqrt{\pi}}}$

3. Originally Posted by redsoxfan325
Close. The derivative is of $f$ is:

$\frac{\tan x\cdot\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot\sec^2x}{\tan^2x}$

Plugging in $\frac{\pi}{4}$ gives us

$\frac{\tan(\pi/4)\cdot\frac{1}{\sqrt{\pi/4}}-2\sqrt{\pi/4}\cdot\sec^2(\pi/4)}{\tan^2(\pi/4)}=\frac{\frac{2}{\sqrt{\pi}}-\sqrt{\pi}\cdot2}{1} = \boxed{\frac{2-2\pi}{\sqrt{\pi}}}$
Ok thanks 1 little error messed me up. Thanks