Calculus 8th Edition

$$y=18x+1$$
First we will find the derivative: $$y'=((3x-1)^{-6})'=-6(3x-1)^{-7}(3x-1)'=-6(3x-1)^{-7}\cdot3=-18(3x-1)^{-7}$$ The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula: $$y-y_{0}=y'(x_{0})(x-x_{0})$$ For $(x_{0},y_{0})=(0,1)$ we get: $$y-1=\left.-18(3x_{0}-1)^{-7}\right|_{x_{0}=0}(x-0)\Rightarrow y=-18(3\cdot0-1)^{-7}\cdot x+1\Rightarrow y=-18\cdot(-1)x+1\Rightarrow y=18x+1$$