If $f ( x ) = \frac { x } { \tan x } , \text { find } f ^ { \prime } \left( \frac { \pi } { 4 } \right)$ .
A) $\frac { 2 - \pi } { 2 }$

B) $\frac { 1 - \pi } { 2 }$
C) $1 - \pi$
D) $\frac { \pi } { 2 }$
E) $1 - 2 \pi$
F) $2 - \pi$

G) $2 - 2 \pi$

H) $- \pi$

Question

If $f ( x ) = \frac { x } { \tan x } , \text { find } f ^ { \prime } \left( \frac { \pi } { 4 } \right)$ .
A) $\frac { 2 - \pi } { 2 }$

B) $\frac { 1 - \pi } { 2 }$
C) $1 - \pi$
D) $\frac { \pi } { 2 }$
E) $1 - 2 \pi$
F) $2 - \pi$

G) $2 - 2 \pi$

H) $- \pi$

Quizplus · Accepted Answer

To find $f'(x)$, we use the quotient rule for differentiation: $$f'(x) = \frac{d}{dx}\left(\frac{x}{\tan(x)}\right) = \frac{\tan(x) \cdot 1 - x \cdot \sec^2(x)}{\tan^2(x)}.$$ At $x = \frac{\pi}{4}$, $\tan\left(\frac{\pi}{4}\right) = 1$ and $\sec^2\left(\frac{\pi}{4}\right) = 2$, so $$f'\left(\frac{\pi}{4}\right) = \frac{1 \cdot 1 - \frac{\pi}{4} \cdot 2}{1^2} = 1 - \frac{\pi}{2} = \frac{2 - \pi}{2}.$$

If f(x)=xtan⁡x, find f′(π4)f ( x ) = \frac { x } { \tan x } , \text { find } f ^ { \prime } \left( \frac { \pi } { 4 } \right)f(x)=tanxx​, find f′(4π​)

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