Question

Is this right? Find the equation of the tangent line to the graph of f(x)=sec(x) at the point (0,1).

Answer 1

$\bf f(x)=sec(x)\implies \left. \cfrac{dy}{dx}=sec(x)tan(x) \right|_{x=0}\implies 0 \\\\\\ \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=0(x-0)\implies y-1=0\implies y=1$