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khush77

25.04.2020

Mathematics

$\displaystyle f'(\frac{\pi}{4})=2$

Step-by-step explanation:

Trigonometric Derivatives

We have the function

$\displaystyle f(x)=\sin x \csc x-\frac{1}{\tan x}$

We must recall that the sine and the cosecant are reciprocal functions, i.e.:

$\displaystyle \sin x =\frac{1}{\csc x}$

Also, the reciprocal of the tangent is the cotangent:

$\displaystyle \cot x =\frac{1}{\tan x}$

Thus:

$\displaystyle f(x)=1-\cot x$

Now it's easier to take the derivative

$\displaystyle f'(x)=0+\csc^2 x= \csc^2 x$

Evaluating for x=π/4:

$\displaystyle f'(\frac{\pi}{4})= \csc^2 \frac{\pi}{4}$

Since csc π/4= $\sqrt{2}$ :

$\displaystyle f'(\frac{\pi}{4})= \sqrt{2}^2$

$\displaystyle f'(\frac{\pi}{4})=2$

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