can someone please help?
This problem asks to find the equation of the tangent line to $y= 3\sin(x)$ at $x= \frac\pi 3$.
I tried finding two points of the graph and finding the slope of the line by taking the derivative of $3\sin(x)$, am I in the right direction? I think I'm doing something wrong because I can't find the correct answer.
Thank you!
Best Answer
1. Find the slope of the line. The slope of the line tangent to $y = 3 \sin(x)$ at $x = \pi/3$ is: $$ \left. \frac{d}{dx}\left[3\sin(x) \right] \right|_{\pi/3} = 3\cos\left(\frac{\pi}{3}\right) = \boxed{\frac{3}{2}}$$ 2. Find a point that the line goes through; since you know that this line is tangent, then there's an easy choice -- $\left(\frac{\pi}{3}, \frac{3\sqrt{3}}{2}\right)$.
3. Use the point-slope formula for a line, namely: $y - y_0 = m(x - x_0)$. Thus, putting it together, you get: $$\boxed{y = \frac{3}{2}\left(x - \frac{\pi}{3}\right) + \frac{3\sqrt{3}}{2}}$$