Answer:
[tex]\displaystyle f'(\frac{\pi}{4})=2[/tex]
Step-by-step explanation:
Trigonometric Derivatives
We have the function
[tex]\displaystyle f(x)=\sin x \csc x-\frac{1}{\tan x}[/tex]
We must recall that the sine and the cosecant are reciprocal functions, i.e.:
[tex]\displaystyle \sin x =\frac{1}{\csc x}[/tex]
Also, the reciprocal of the tangent is the cotangent:
[tex]\displaystyle \cot x =\frac{1}{\tan x}[/tex]
Thus:
[tex]\displaystyle f(x)=1-\cot x[/tex]
Now it's easier to take the derivative
[tex]\displaystyle f'(x)=0+\csc^2 x= \csc^2 x[/tex]
Evaluating for x=π/4:
[tex]\displaystyle f'(\frac{\pi}{4})= \csc^2 \frac{\pi}{4}[/tex]
Since csc π/4=[tex]\sqrt{2}[/tex]:
[tex]\displaystyle f'(\frac{\pi}{4})= \sqrt{2}^2[/tex]
[tex]\displaystyle f'(\frac{\pi}{4})=2[/tex]
