If $f(x)=\dfrac{x}{\tan{x}}$, then $f'(\frac{\pi}{4})=$

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Using the quotient rule,

$$f(x)=\frac{x}{\tan{x}}$$ $$f'(x)=\frac{\tan{x}(1)-x(\sec^2{x})}{\tan^2{x}}$$ $$= \frac{\tan{x}-x\sec^2{x}}{\tan^2{x}}$$

Using our knowledge of the unit circle,

$$\tan{\frac{\pi}{4}}=1$$ $$\sec{\frac{\pi}{4}}=\frac{1}{\cos{\frac{\pi}{4}}} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$$ $$\sec^2{\left(\frac{\pi}{4}\right)}=\left(\frac{2}{\sqrt2}\right)^2=\frac{4}{2}=2$$

Substituting our values,

$$f'(x) = \frac{\tan{x}-x\sec^2{x}}{\tan^2{x}}$$ $$f'\left(\frac{\pi}{4}\right) = \frac{\tan{(\frac{\pi}{4})}-\frac{\pi}{4}\sec^2{(\frac{\pi}{4}})}{\tan^2{\left(\frac{\pi}{4}\right)}}$$ $$= \frac{1-\frac{\pi}{4}(2)}{(1)^2}$$ $$= \boxed{1-\frac{\pi}{2}}$$