Find an equation of the tangent line to the curve at the given point. y = sin(3x) sin2 (3x) given the point (0,0) Find an equation of the tangent line to the curve $y = sin(3x) + sin^2 (3x)$  given the point (0,0). Answer is $y = 3x$, but please explain solution steps. 
 A: Hint:
Do you know that the slope of the tangent line at a point of the graph of a function is the derivative of the function at this point?
So, for $y = \sin(3x) + \sin^2 (3x)$  find the derivative $y'$ ( can you do?), then evaluate this derivative  for $x=0$ 
Now the line has equation $y=mx$ with $m=y'(0)$

Using the chain rule the derivative is:
$$
y'=\cos(3x)\cdot(3x)'+2\sin(3x)(\sin(3x))'=3\cos(3x)+2\sin(3x)\cos(3x)(3x)'$$$$=3\cos(3x)+6\sin(3x)\cos(3x)
$$
so $y'(0)=3$.
A: Goal
Find the slope $m$, and intercept $b$, for the line
$$
 y = mx + b,
$$
tangent at the origin to the curve
$$
 f(x) = \sin ^2(3 x)+\sin (3 x).
$$
Intercept $b$
Because the function goes through the origin, the tangent line will also go through the origin. Therefore the $y-$intercept $b=0$.
Slope $m$
The slope of the tangent line $m$ is, by definition, the same as the slope of the target function at the point of contact. The slope of the function is
$$
  f'(x) = 6 \sin (3 x) \cos (3 x) + 3 \cos (3 x).
$$
The problem specifies $x_{*} = 0.$ The point of contact is
$$
  \left( x_{*}, f(x_{*}) \right) = \left( 0, 3 \right).
$$
The slope of the function at the point of contact is 
$$
  f'( x_{*} ) = 6 \sin (3 x_{*}) \cos (3 x_{*}) + 3 \cos (3 x_{*}) = 6\cdot 0 \cdot 0 + 3 \cdot 1 = 3.
$$
The slope of the tangent line is
$$
  m = f'( x_{*} ) = 3.
$$
Solution
The equation of the line tangent to $f(x)$ at $x_{*} = 0$ is
$$
  y = mx + b = 3x.
$$

