Answer
$y=-6x+8$
Work Step by Step
$f(x)=\frac{2}{\sqrt[4] x^{3}}$
$f(x)=2x^{-\frac{3}{4}}$
$f'(x)=-6x^{-\frac{7}{4}}$
$f'(1)=-6(1)^{-\frac{7}{4}}$
$f'(1)=-6$
use equation of a straight line $y-y_{1}=m(x-x_{1})$
$y-2=-6(x-1)$
$y-2=-6x+6$
$y=-6x+8$
