Question

Find an equation of the tangent line to y = 2e^x sec(x) at x = pi/4

Answer 1

Answer:

y = 6.20 + 12.4(x - pi/4)

Step-by-step explanation:

dy/dx = 2e^x secx + 2e^x secx * tanx= 2e^x secx (1 + tanx)

At x = pi/4, use sec(pi/4) = $\sqrt{2}$. tan(pi/4) =1

so dy/dx = 2e^(pi/4) * $\sqrt{2}$ * (1 + 1) = $4 \sqrt{2} e^{\pi/4}$ ≈ 12.4 (This is the slope of the tangent line)

When x = pi/4, y =  2e^(pi/4) * $\sqrt{2}$  = $2 \sqrt{2} e^{\pi/4}$= 6.20

Then the equation of the tangent line is

y = 6.20 + 12.4(x - pi/4)