find the equation of the tangent line to the curve ( y = 4sec(x)-8cos(x) ) at the point ( (pi/3,4) ). write…

Explanation:

Step1: Find the derivative of the function

The derivative of (y = 4\sec(x)-8\cos(x)) is (y^\prime=4\sec(x)\tan(x)+8\sin(x)).

Step2: Evaluate the derivative at (x = \frac{\pi}{3})

Substitute (x=\frac{\pi}{3}) into (y^\prime): (\sec(\frac{\pi}{3}) = 2), (\tan(\frac{\pi}{3})=\sqrt{3}), (\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}) (y^\prime|_{x = \frac{\pi}{3}}=4\times2\times\sqrt{3}+8\times\frac{\sqrt{3}}{2}=8\sqrt{3}+4\sqrt{3}=12\sqrt{3}) So the slope (m = 12\sqrt{3})

Step3: Use the point - slope form (y - y_1=m(x - x_1))

We have (x_1=\frac{\pi}{3}), (y_1 = 4) (y-4=12\sqrt{3}(x-\frac{\pi}{3})) (y=12\sqrt{3}x-4\pi\sqrt{3}+4)

Answer:

(y = 12\sqrt{3}x+(4 - 4\pi\sqrt{3}))