find an equation of the tangent line to the curve at the point (3π, 0). y = sin(sin(x)) y =

Explanation:

Step1: Find the derivative using chain - rule

Let $u = \sin(x)$, then $y=\sin(u)$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=\cos(u)$ and $\frac{du}{dx}=\cos(x)$. So $\frac{dy}{dx}=\cos(\sin(x))\cdot\cos(x)$.

Step2: Evaluate the derivative at $x = 3\pi$

Substitute $x = 3\pi$ into $\frac{dy}{dx}$. $\cos(\sin(3\pi))\cdot\cos(3\pi)$. Since $\sin(3\pi)=0$, then $\cos(\sin(3\pi))=\cos(0) = 1$, and $\cos(3\pi)=- 1$. So $\frac{dy}{dx}\big|_{x = 3\pi}=1\times(-1)=-1$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(3\pi,0)$ and $m=-1$. $y-0=-1(x - 3\pi)$.

Answer:

$y=-x + 3\pi$