find the equation of the tangent line to the curve ( y = 6sin x ) at the point ( left( \frac{pi}{6},3
ight) ).
the equation of this tangent line can be written in the form ( y = mx + b ) where
( m=)
and ( b=)

Question

Accepted Answer

# Answer:
$m = 3\sqrt{3}$, $b = 3-\frac{\pi\sqrt{3}}{2}$

# Explanation:
## Step1: Find the derivative of the function
The derivative of $y = 6\sin x$ is $y^\prime=6\cos x$.
## Step2: Calculate the slope $m$
Substitute $x = \frac{\pi}{6}$ into $y^\prime$.
$$
\begin{align*}
m&=6\cos\frac{\pi}{6}\
&=6\times\frac{\sqrt{3}}{2}\
&=3\sqrt{3}
\end{align*}
$$
## Step3: Find the $y -$intercept $b$
We know the equation of the line is $y=mx + b$, and the line passes through the point $(\frac{\pi}{6},3)$. Substitute $x=\frac{\pi}{6}$, $y = 3$ and $m = 3\sqrt{3}$ into $y=mx + b$.
$$
\begin{align*}
3&=3\sqrt{3}\times\frac{\pi}{6}+b\
3&=\frac{\pi\sqrt{3}}{2}+b\
b&=3-\frac{\pi\sqrt{3}}{2}
\end{align*}
$$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the derivative of the function

Step2: Calculate the slope \(m\)

Step3: Find the \(y -\)intercept \(b\)

Answer: