find the equation of the tangent line to the curve ( y = 4sec(x)-8cos(x) ) at the point ( (pi/3,4) ). write your answer in the form ( y = mx + b ) where ( m ) is the slope and ( b ) is the ( y )-intercept

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Accepted Answer

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

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QUESTION IMAGE

Question

Explanation:

Step1: Find the derivative of the function

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

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

Answer: