   Chapter 3.4, Problem 39E

Chapter
Section
Textbook Problem

Find the derivative of the function.f(t) = tan(sec(cos t))

To determine

To find:  The derivative of f(t)=tan(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=fg 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)=nxn1 (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)

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