## Calculus 10th Edition

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