#### To determine

**To find:** The rate of change of the brightness after *t* days.

#### Answer

The rate of change of the brightness after *t* days is B′(t)=7π54cos(2πt5.4)_.

#### Explanation

**Given:**

The function B(t)=4.0+0.35sin(2πt5.4).

**Derivative rule:**

(1) *Constant Multiple Rule:* ddx[cf(x)]=cddx[f(x)]

(2) *Sum Rule:* ddx(f+g)=ddx(f)+ddx(g)

(3) *Quotient Rule:* ddx(fg)=gddx(f)−fddx(g)(g)2

**Result used:** *Chain Rule*

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

F′(x)=f′(g(x))⋅g′(x) (1)

**Calculation:**

Obtain the derivative of B(t).

B′(t)=ddt(B(t))=ddt(4.0+0.35sin(2πt5.4))

Apply the sum rule (1) and the constant multiple rule (2),

B′(t)=ddt(4.0)+ddt(0.35sin(2πt5.4))=0+0.35ddt(sin(2πt5.4))

B′(t)=0.35ddt(sin(2πt5.4)) (2)

Obtain the derivative ddt(sin(2πt5.4)) by using the chain rule as shown in equation (1),

Let g(t)=2πt5.4 and f(u)=sinu where u=g(t).

ddt(sin(2πt5.4))=f′(g(t))⋅g′(t) (3)

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

f′(g(t))=f′(u)=ddu(f(u))=ddu(sinu)=cosu

Substitute u=2πt5.4 in the above equation,

f′(g(t))=cos(2πt5.4)

Thus, the derivative is f′(g(t))=cos(2πt5.4).

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

g′(t)=ddt(2πt5.4)=2π5.4ddt(t)=2π5.4(1)=2π5.4

Thus, the derivative is g′(t)=2π5.4.

Substitute cos(2πt5.4) for f′(g(t)) and 2π5.4 for g′(t) in equation (3),

ddt(sin(2πt5.4))=cos(2πt5.4)(2π5.4)=2π5.4cos(2πt5.4)

Substitute 2π5.4cos(2πt5.4) for ddt(sin(2πt5.4)) in equation (2),

B′(t)=0.35(2π5.4cos(2πt5.4))=0.7π5.4cos(2πt5.4)≈0.407cos(2πt5.4)

Therefore, the rate of change of the brightness after *t* days is B′(t)=0.41cos(2πt5.4)_.

#### To determine

**To find:** The rate of increase after one day and correct to two decimal places.

#### Answer

The rate of increase after one day is B′(1)=0.16.

#### Explanation

**Calculation:**

From part (a), the rate of change of the brightness after *t* days is B′(t)=0.41cos(2πt5.4).

Substitute t=1 in the above equation,

B′(1)=7π54cos(2π(1)5.4)=7π54cos(2π5.4)≈(0.407)(0.396)=0.161172

Therefore, the rate of increase after one day is B′(1)=0.16.