Question

find an equation of the tangent line to the curve at the given point. y = sin(sin x), (4π, 0)

Explanation:

Step1: Compute derivative using chain rule

Let $u = \sin x$, so $y = \sin u$. Then $\frac{dy}{du} = \cos u$ and $\frac{du}{dx} = \cos x$. Thus $\frac{dy}{dx} = \cos(\sin x) \cdot \cos x$.

Step2: Find slope at $x=4\pi$

$\cos(4\pi) = 1$, $\sin(4\pi) = 0$, so $\frac{dy}{dx}\big|_{x=4\pi} = \cos(0) \cdot 1 = 1$.

Step3: Write tangent line equation

Using point-slope form $y - y_1 = m(x - x_1)$: $y - 0 = 1 \cdot (x - 4\pi)$.

Answer:

$y = x - 4\pi$