#### To determine

**To find:** The derivative of f(t)=tan(sec(cost)).

#### Answer

The derivative of the function f(t)=tan(sec(cost)) is f′(t)=−sint⋅sec(cost)⋅tan(cost)⋅sec2(sec(cost))_.

#### Explanation

**Given:**

The function is f(t)=tan(sec(cost)).

**Formula used:**

**The Chain Rule:**

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

F′(t)=f′(g(t))⋅g′(t) (1)

**Power Rule:**

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

**Calculation:**

Obtain the derivative of f(t).

f′(t)=ddt(f(t))=ddt(tan(sec(cost)))

Let h(t)=sec(cost) and g(u)=tanu where u=h(t).

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

f′(t)=g′(h(t))⋅h′(t) (3)

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

g′(h(t))=f′(u)=ddu(f(u))=ddu(tanu)=sec2u

Substitute u=sec(cost) in the above equation,

g′(h(t))=sec2(sec(cost))

Thus, the derivative is g′(h(t))=sec2(sec(cost)).

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

h′(t)=ddt(sec(cost))

Let k(t)=cost and s(u)=secu where u=k(t).

Apply the chain rule as shown in equation (1)

h′(t)=s′(k(t))⋅k′(t) (4)

The derivative s′(k(t)) is computed as follows,

s′(k(t))=f′(u)=ddu(f(u))=ddu(secu)=secutanu

Substitute u=cost in the above equation,

f′(k(t))=sec(cost)tan(cost)

Thus, the derivative is f′(k(t))=sec(cost)tan(cost).

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

k′(t)=ddt(cost)=−sint

Thus, the derivative of k(t) is k′(t)=−sint.

Substitute sec(cost)tan(cost) for f′(k(t)) and −sint for k′(t) in equation (4),

s′(k(t))⋅k′(t)=sec(cost)tan(cost)(−sint)=−sintsec(cost)tan(cost)

Thus, the derivative of h(t) is h′(t)=−sintsec(cost)tan(cost).

Substitute sec2(sec(cost)) for g′(h(t)) and −sintsec(cost)tan(cost) for h′(t) in equation (3),

g′(h(t))⋅h′(t)=[sec2(sec(cost))]⋅[−sint⋅sec(cost)⋅tan(cost)]=−sint⋅sec(cost)⋅tan(cost)⋅sec2(sec(cost))

Therefore, the derivative of f(t)=tan(sec(cost)) is f′(x)=−sint⋅sec(cost)⋅tan(cost)⋅sec2(sec(cost))_.