17. [-/1 Points] DETAILS MY NOTES SCALCET9 3.XP.4.051. Find an equation of the tangent line to the curve at the given point. $$ y=\sin (9 x)+\sin ^{2}(9 x),(0,0) $$ F

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1. We are given the curve  
   $$
   y = \sin(9x) + \sin^2(9x)
   $$
   and the point $$(0,0)$$.

2. First, compute the derivative $$ y' $$ to find the slope of the tangent line.

3. Differentiate term by term.

- For the term $$\sin(9x)$$, use the chain rule:
     $$
     \frac{d}{dx} [\sin(9x)] = \cos(9x) \cdot 9.
     $$
     
   - For the term $$\sin^2(9x)$$, write it as $$(\sin(9x))^2$$ and differentiate using the chain rule:
     $$
     \frac{d}{dx} [\sin^2(9x)] = 2\sin(9x) \cdot \frac{d}{dx}[\sin(9x)] = 2\sin(9x) \cdot \cos(9x) \cdot 9.
     $$
     This simplifies to:
     $$
     18\sin(9x)\cos(9x).
     $$

4. Combine both parts to get the derivative:
   $$
   y' = 9\cos(9x) + 18\sin(9x)\cos(9x).
   $$

5. Evaluate the derivative at $$ x = 0 $$.

- At $$ x = 0 $$:
     $$
     \sin(9 \cdot 0) = \sin(0) = 0, \quad \cos(9 \cdot 0) = \cos(0) = 1.
     $$
   - Thus:
     $$
     y'(0) = 9\cdot1 + 18\cdot0\cdot1 = 9.
     $$
     
6. The slope of the tangent line at $$(0,0)$$ is $$ 9 $$.

7. Use the point-slope form of a line:
   $$
   y - y_1 = m(x - x_1)
   $$
   where $$ m = 9 $$ and the point $$(x_1, y_1) = (0,0) $$. Plug in these values:
   $$
   y - 0 = 9(x - 0)
   $$
   Simplify:
   $$
   y = 9x.
   $$

The equation of the tangent line to the curve at the point $$(0,0)$$ is  
$$
y = 9x.
$$
The equation of the tangent line at $$(0,0)$$ is $$ y = 9x $$.

[-/1 Points] DETAILS MY NOTES SCALCET9 3.XP.4.051.
Find an equation of the tangent line to the curve at the given point.

F

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[-/1 Points] DETAILS MY NOTES SCALCET9 3.XP.4.051. Find an equation of the tangent line to the curve at the given point. F