Answer
$y=4x+1-\pi.$
Work Step by Step
Using the Chain Rule:
$u=\tan{x}$; $\dfrac{du}{dx}=\sec^2{x}$
$\dfrac{d}{du}f(u)=2u$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=2\sec^2{x}\tan{x}.$
$f'(\dfrac{\pi}{4})=2\sec^2{\dfrac{\pi}{4}\tan{\dfrac{\pi}{4}}}=4.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-1)=4(x-\dfrac{\pi}{4})\rightarrow y=4x+1-\pi.$
A graphing calculator and a computer algebra system have been used to confirm these results.
