To determine

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

Answer

The first derivative of y=1(1+tanx)2 is y′=−2sec2x(1+tanx)3_

The second derivative of y=1(1+tanx)2 is y″=2sec2x(3tan2x−2tanx+3)(1+tanx)4_

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=g∘h 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)=−2u−3

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)−3⋅sec2x)

    =−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.

Substitute 2secx for s′(k(x)) and secxtanx for h′(x) in equation (5),

ddx(sec2x)=2secx(secxtanx)=2sec2xtanx

Thus, the derivative is ddx(sec2x)=2sec2xtanx.

Obtain the derivative ddx(−2(1+tanx)−3) by using the chain rule (1),

Let p(x)=1+tanx and f(u)=−2u−3  where u=p(x)

ddx(−2(1+tanx)−3)=f′(p(x))⋅p′(x) (6)

The derivative f′(p(x)) is computed as follows,

f′(p(x))=f′(u)=ddu(f(u))=ddu(−2u)−3=−2(−3u−3−1)

Substitute u=1+tanx in the above equation,

f′(p(x))=6(1+tanx)−3−1=6(1+tanx)−4=6(1+tanx)4

Thus the derivative is f′(p(x))=6(1+tanx)4.

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

p′(x)=ddx(1+tanx)=ddx(1)+ddx(tanx)=(0)+sec2x=sec2x

Thus, the derivative is p′(x)=sec2x.

Substitute 6(1+tanx)4 for f′(p(x)) and sec2x for p′(x) in equation (6),

ddx(−2(1+tanx)−3)=6(1+tanx)4(sec2x)=6sec2x(1+tanx)4

Thus, the derivative of is ddx(−2(1+tanx)−3)=6sec2x(1+tanx)4_.

Substitute 2sec2xtanx for ddx(sec2x) and 6sec2x(1+tanx)4 for ddx(−2(1+tanx)−3) in equation (4)

y″=−2(1+tanx)−3(2sec2xtanx)+sec2x(6sec2x(1+tanx)4)=−4sec2xtanx(1+tanx)3+6sec2xsec2x(1+tanx)4=2sec2x(−2(1+tanx)tanx+3sec2x)(1+tanx)4=2sec2x((−2tanx−2tan2x)+3(tan2x+1))(1+tanx)4  (Qsec2x=tan2x+1 )

Simplify further and obtain the derivative,

y″=2sec2x(−2tanx−2tan2x+3tan2x+3)(1+tanx)4=2sec2x(tan2x−2tanx+3)(1+tanx)4

Therefore, the derivative of y is y″=2sec2x(tan2x−2tanx+3)(1+tanx)4_.