#### To determine

**To find:** The derivative of the function.

#### Answer

The derivative of y=cot2(sinθ) is y′=− 2cot(sinθ)csc2(sinθ)⋅cosθ.

#### Explanation

**Given:**

The function is y=cot2(sinθ).

**Formula used:**

**The Chain Rule:**

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

F′(θ)=f′(g(θ))⋅g′(θ) (1)

**Power Rule:**

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

**Calculation:**

Obtain the derivative of y.

y′=ddx(y)=ddx(cot2(sinθ))

Let h(θ)=cot(sinθ) and f(u)=u2 where u=h(θ)

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

y′=f′(h(θ))⋅h′(θ) (3)

The derivative f′(h(θ)) is computed as follows,

f′(h(θ))=f′(u)=ddu(f(u))=ddu(u2)

Apply the power rule as shown in equation (2) then substitute u=cot(sinθ),

f′(h(θ))=(2u2−1)=2u=2cot(sinθ)

Thus, the derivative is f′(h(θ))=2cot(sinθ).

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

h′(θ)=ddθ(cot(sinθ)).

Let k(θ)=sinθ and g(u)=cotu where u=k(θ).

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

h′(θ)=g′(k(θ))⋅k′(θ) (4)

The derivative g′(k(θ)) is computed as follows,

g′(k(θ))=g′(u)=ddu(g(u))=ddu(cotu)=−csc2u

Substitute u=sinθ in above equation,

f′(k(θ))=−csc2(sinθ).

The derivative of k(θ) is computed as follows,

k′(θ)=ddθ(sinθ)=cosθ

Thus, the derivative of k(θ) is k′(θ)=cosθ.

Substitute −csc2(sinθ) for f′(k(θ)) and cosθ for k′(θ) in equation (4),

g′(k(θ))⋅k′(θ)=−csc2(sinθ)(cosθ)=−(cosθ)csc2(sinθ)

Thus, the derivative of h(θ) is h′(θ)=−(cosθ)csc2(sinθ).

Substitute 2cot(sinθ) for f′(h(θ)) and −(cosθ)csc2(sinθ) for h′(θ) in equation (3),

f′(h(θ))⋅h′(θ)=[ 2cot(sinθ)][−csc2(sinθ)(cosθ)]=− 2cot(sinθ)csc2(sinθ)⋅cosθ

Therefore, the derivative of y=cot2(sinθ) is y′=− 2cot(sinθ)csc2(sinθ)⋅cosθ_.