Instantaneous rate of change [Trig]

[anonymous12]

Homework Statement

For the function f(x) = 3sin 2x, find the instantaneous rate of change at point A (\frac{\pi}{6}, f(\frac{\pi}{6}))

Homework Equations

iroc = instantaneous rate of change
iroc= \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}

The Attempt at a Solution

iroc = instantaneous rate of change
iroc= \lim_{h\to 0} \frac{(3sin2(\frac{\pi}{6}+0.001)) - (3sin2(\frac{\pi}{6})}{0.001}
iroc= \lim_{h\to 0} \frac{(3sin(\frac{\pi}{3}+0.002)) - (3sin(\frac{\pi}{3}))}{0.001}
iroc= \lim_{h\to 0} \frac{(3(\frac{\sqrt3}{2}+0.002)) - (3(\frac{\sqrt{3}}{2}))}{0.001}
iroc= \lim_{h\to 0} \frac{(3(\frac{\sqrt3 + 0.004}{2})) - (\frac{3\cdot\sqrt{3}}{2}))}{0.001}
iroc= \lim_{h\to 0} \frac{(\frac{3\sqrt3 + 0.012}{2})) - (\frac{3\sqrt{3}}{2}))}{0.001}
iroc = 6

Is this answer correct?

 

[6.28318531]

You have ignored the limit, if you want to do it that way you have to vary h, so start with h=0.001 and then go down to h=0.000000001 for instance. Then once you have collected all the results you can see what it limits to, but for here just try a really small h. And you also seem to be off by a factor of 1/2, you have to take the sin of everything in the brackets.