#### To determine

**To find:** The composite function in the form f(g(x)) and obtain the derivative of *y*.

#### Answer

The inner function is u=sinx and the outer function is f(u)=u.

The derivative of *y* is y′(x)=cosx2sinx_.

#### Explanation

**Given:**

The function is y=sinx.

**Formula used:**

**The 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)

**Power Rule:**

If *n* is positive integer, then ddx(xn)=nxn−1 (2)

**Calculation:**

Let the inner function be u=g(x) and the outer function be y=f(u).

Then, g(x)=sinx and f(u)=u. That is,

y=sinx=f(sinx)=f(g(x))

Therefore, y=f(g(x)).

Hence, the inner function is u=sinx and the outer function is f(u)=u.

Thus, the required form of composite function is f(g(x))=u.

Obtain the derivative of *y* as follows.

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

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

y′(x)=g′(h(x))⋅h′(x) (3)

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

g′(h(x))=g′(u)=ddu(g(u))=dduu=12u

Substitute u=sinx in above equation,

g′(h(x))=12sinx

Thus, the derivative g′(h(x)) is g′(h(x))=12sinx.

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 12sinx for g′(h(x)) and cosx for h′(x) in equation (3),

g′(h(x))⋅h′(x)=12sinx(cosx)=cosx2sinx

Therefore, the derivative of y is y′(x)=cosx2sinx_.