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'=4\sec(x)\tan(x)+8\sin(x)) (using the derivatives (\frac{d}{dx}\sec(x)=\sec(x)\tan(x)) and (\frac{d}{dx}\cos(x)=-\sin(x))).

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

When (x=\frac{\pi}{3}), (\sec(\frac{\pi}{3}) = 2), (\tan(\frac{\pi}{3})=\sqrt{3}), (\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}). [ \begin{align*} y'(\frac{\pi}{3})&=4\times2\times\sqrt{3}+8\times\frac{\sqrt{3}}{2}\ &=8\sqrt{3}+4\sqrt{3}\ &=12\sqrt{3} \end{align*} ] 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) and (m = 12\sqrt{3}). [ \begin{align*} y-4&=12\sqrt{3}(x-\frac{\pi}{3})\ y-4&=12\sqrt{3}x- 4\pi\sqrt{3}\ y&=12\sqrt{3}x+(4 - 4\pi\sqrt{3}) \end{align*} ]

Answer:

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