find the equation of the tangent line to the curve ( y = 6sin x ) at the point ( left( \frac{pi}{6},3…

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*} ]