Question

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

I took the derivative of secx and got secxtanx. Then, I plugged in 0 for x and got 1x0, which is 0.

y=0

I think it's wrong, but I don't understand why.

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`