Answer :
To find the equation of the tangent line to the curve [tex]\( y = \sin(\sin(x)) \)[/tex] at the point [tex]\( (3\pi, 0) \)[/tex], we need to follow these steps:
1. Find the derivative of the function [tex]\( y = \sin(\sin(x)) \)[/tex]:
- The derivative [tex]\( \frac{dy}{dx} \)[/tex] will give us the slope of the tangent line.
- Since [tex]\( y \)[/tex] is a composition of functions, we apply the chain rule.
- Let [tex]\( u = \sin(x) \)[/tex], then [tex]\( y = \sin(u) \)[/tex].
- The chain rule states [tex]\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)[/tex].
[tex]\[ \frac{dy}{du} = \cos(u) = \cos(\sin(x)) \quad \text{and} \quad \frac{du}{dx} = \cos(x) \][/tex]
[tex]\[ \frac{dy}{dx} = \cos(\sin(x)) \cdot \cos(x) \][/tex]
2. Evaluate the derivative at the given point [tex]\( x = 3\pi \)[/tex]:
[tex]\[ \frac{dy}{dx} \bigg|_{x = 3\pi} = \cos(\sin(3\pi)) \cdot \cos(3\pi) \][/tex]
- Note [tex]\( \sin(3\pi) = 0 \)[/tex], so [tex]\( \cos(\sin(3\pi)) = \cos(0) = 1 \)[/tex].
- And [tex]\( \cos(3\pi) = -1 \)[/tex].
[tex]\[ \frac{dy}{dx} \bigg|_{x = 3\pi} = 1 \cdot (-1) = -1 \][/tex]
Thus, the slope of the tangent line at [tex]\( (3\pi, 0) \)[/tex] is [tex]\( -1 \)[/tex].
3. Use the point-slope form of the equation of the tangent line:
- The point-slope form is given by [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is the point on the curve.
- Here, [tex]\( m = -1 \)[/tex], [tex]\( x_1 = 3\pi \)[/tex], and [tex]\( y_1 = 0 \)[/tex].
[tex]\[ y - 0 = -1(x - 3\pi) \][/tex]
Simplifying this, we get:
[tex]\[ y = -x + 3\pi \][/tex]
Therefore, the equation of the tangent line to the curve [tex]\( y = \sin(\sin(x)) \)[/tex] at the point [tex]\((3\pi, 0)\)[/tex] is:
[tex]\[ \boxed{y = -x + 3\pi} \][/tex]
1. Find the derivative of the function [tex]\( y = \sin(\sin(x)) \)[/tex]:
- The derivative [tex]\( \frac{dy}{dx} \)[/tex] will give us the slope of the tangent line.
- Since [tex]\( y \)[/tex] is a composition of functions, we apply the chain rule.
- Let [tex]\( u = \sin(x) \)[/tex], then [tex]\( y = \sin(u) \)[/tex].
- The chain rule states [tex]\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)[/tex].
[tex]\[ \frac{dy}{du} = \cos(u) = \cos(\sin(x)) \quad \text{and} \quad \frac{du}{dx} = \cos(x) \][/tex]
[tex]\[ \frac{dy}{dx} = \cos(\sin(x)) \cdot \cos(x) \][/tex]
2. Evaluate the derivative at the given point [tex]\( x = 3\pi \)[/tex]:
[tex]\[ \frac{dy}{dx} \bigg|_{x = 3\pi} = \cos(\sin(3\pi)) \cdot \cos(3\pi) \][/tex]
- Note [tex]\( \sin(3\pi) = 0 \)[/tex], so [tex]\( \cos(\sin(3\pi)) = \cos(0) = 1 \)[/tex].
- And [tex]\( \cos(3\pi) = -1 \)[/tex].
[tex]\[ \frac{dy}{dx} \bigg|_{x = 3\pi} = 1 \cdot (-1) = -1 \][/tex]
Thus, the slope of the tangent line at [tex]\( (3\pi, 0) \)[/tex] is [tex]\( -1 \)[/tex].
3. Use the point-slope form of the equation of the tangent line:
- The point-slope form is given by [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is the point on the curve.
- Here, [tex]\( m = -1 \)[/tex], [tex]\( x_1 = 3\pi \)[/tex], and [tex]\( y_1 = 0 \)[/tex].
[tex]\[ y - 0 = -1(x - 3\pi) \][/tex]
Simplifying this, we get:
[tex]\[ y = -x + 3\pi \][/tex]
Therefore, the equation of the tangent line to the curve [tex]\( y = \sin(\sin(x)) \)[/tex] at the point [tex]\((3\pi, 0)\)[/tex] is:
[tex]\[ \boxed{y = -x + 3\pi} \][/tex]
