#### To determine

**To find:** The derivative of J(θ)=tan2(nθ).

#### Answer

The derivative of J(θ)=tan2(nθ) is J′(θ)=2ntannθsec2nθ_.

#### Explanation

**Given:**

The function is J(θ)=tan2(nθ).

**Result 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′(h(θ))⋅h′(θ) (1)

**Power Rule:**

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

**Calculation:**

Obtain the derivative of J(θ).

J′(θ)=ddθ(J(θ))=ddθ(tan2(nθ))

Let h(θ)=tannθ and f(u)=u2 where u=h(θ).

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

J′(θ)=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) the substitute u=tannθ,

f′(h(θ))=(2u2−1)=2u

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

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

h′(θ)=ddθ(tannθ)=sec2nθ⋅n=nsec2nθ

Thus, the derivative of h(θ) is h′(θ)=nsec2nθ.

Substitute 2tannθ for f′(h(θ)) and nsec2nθ for h′(θ) in equation (3),

f′(h(θ))⋅h′(θ)=(2tannθ)(nsec2nθ)=2ntan(nθ)sec2(nθ)

Therefore, the derivative of J(θ)=tan2(nθ) is J′(θ)=2ntan(nθ)sec2(nθ)_.