   Chapter 3.4, Problem 48E

Chapter
Section
Textbook Problem

Find y′ and y″. y = 1 ( 1 + tan x ) 2

To determine

To find: The first and second derivatives of y. That is,y and y.

Explanation

Given:

The function is y=1(1+tanx)2.

Result used: Chain Rule:

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh 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)

Product Rule:

If f(x). and g(x) are both differentiable function, then

ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)] (2)

Calculation:

Obtain the first derivative of

y=ddx(y)=ddx(1(1+tanx)2)=ddx((1+tanx)2)

Let h(x)=1+tanx and g(u)=(u)2  where u=h(x).

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

y=g(h(x))h(x) (3)

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

g(h(x))=g(u)=ddu(g(u))=ddu((u)2)=2u3

Substitute u=1+tanx in the above equation,

g(h(x))=2(1+tanx)3

Thus, the derivative is g(h(x))=2(1+tanx)3.

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

h(x)=ddx(1+tanx)=ddx(1)+ddx(tanx)=(0)+sec2x=sec2x

Thus the derivative is h(x)=sec2x.

Substitute 2(1+tanx)3 for g(h(x)) and sec2x for h(x) in equation (3),

y=2(1+tanx)3(sec2x)=2sec2x(1+tanx)3=2sec2x(1+tanx)3

Therefore, the derivative of y=1(1+tanx)2 is y=2sec2x(1+tanx)3_.

Obtain the second derivative of y.

y=ddx(y)=ddx(2sec2x(1+tanx)3)

Apply the product rule as shown in equation (2),

y=ddx(2(1+tanx)3sec2x)

=2(1+tanx)3ddx(sec2x)+sec2xddx(2(1+tanx)3) (4)

Obtain the derivative ddx(sec2x) by using the chain rule as shown in equation (1),

Let k(x)=secx and s(u)=u2  where u=k(x).

ddx(sec2x)=s(k(x))k(x) (5)

The derivative s(k(x)) is computed as follows,

s(k(x))=s(u)=ddu(s(u))=ddu((u)2)=2u

Substitute u=secx in the above equation,

s(k(x))=2secx

Thus, the derivative is s(k(x))=2secx.

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

k(x)=ddx(secx)=secxtanx

Thus, the derivative is k(x)=secxtanx

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