#### To determine

**To find:** The equation of the tangent line to the curve at the point.

#### Answer

The equation of the tangent line to curve y=sin(sinx) at (π,0) is y=−x+π_.

#### Explanation

**Given:**

The function is y=sin(sinx).

The point is (π,0).

**Result used:** Chain Rule

If *h* is differentiable at *x* and *g* is differentiable at h(x), then the composite function F=g∘h defined by F(x)=g(h(x)) is differentiable at *x* and F′ is given by the product,

F′(x)=g′(h(x))⋅h′(x) (1)

**Formula used:**

The equation of the tangent line at (x1,y1) is y−y1=m(x−x1) (2)

Where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y)=ddx(sin(sinx))

Let h(x)=sinx and g(u)=sinu where u=h(x).

Apply the chain rule as shown in equation (1),

dydx=g′(h(x))⋅h′(x) (3)

The derivative g′(h(x)) is computed as follows,

g′(h(x))=g′(u)=ddu(g(u))=ddu(sinu)=cosu

Substitute u=sinx in above equation,

g′(h(x))=cos(sinx)

Thus, the derivative is g′(h(x))=cos(sinx).

The derivative of h(x) is computed as follows,

h′(x)=ddx(sinx)=cosx

Thus, the derivative of h(x) is h′(x)=cosx.

Substitute cos(sinx) for g′(h(x)) and cosx for h′(x) in equation (3),

dydx=cos(sinx)(cosx)=(cosx)(cos(sinx))

Therefore, the derivative of y=sin(sinx) is dydx=(cosx)(cos(sinx))_.

The slope of the tangent line at (π,0) is computed as follows,

m=dydx|x=π=(cosπ)(cos(sinπ)) =(−1)(cos(0)) =−(1)=−1

Thus, the slope of the tangent line at (π,0) is m=−1 .

Substitute (π,0) for (x1,y1) and m=−1 in equation (2),

y−0=−1(x−π)y=−x+π

Therefore, the equation of the tangent line to curve y=sin(sinx) at (π,0) is y=−x+π_.