#### To determine

**To find:** The derivative of the function f(x)=x4+2x.

#### Answer

The derivative of f(x) is f′(x)=4x3+2.

#### Explanation

**Given:**

The function is f(x)=x4+2x.

**Formula used:**

The derivative of a function *f ,* denoted by f′(x), is

f′(x)=limh→0f(x+h)−f(x)h (1)

**Calculation:**

Obtain the derivative of the function f(x).

Compute f′(x) by using the equation (1),

f′(x)=limh→0f(x+h)−f(x)h=limh→0(x+h)4+2(x+h)−(x4+2x)h

Expand the numerator,

f′(x)=limh→0(x4 + 4x3h + 6x2h2 +4xh3 +h4)+2x+2h−x4−2xh=limh→0 4x3h+6x2h2+4xh3+h4+2hh=limh→0h( 4x3+6x2h+4xh2+h3+2)h

Since the limit *h* approaches zero but not equal to zero, cancel the common term *h* from both the numerator and the denominator,

f′(x)=limh→0(4x3+6x2h+4xh2+h3+2)=4x3+6x2(0)+4x(0)2 +(0)3+2=4x3+2

Thus, the value of the derivative is f′(x)=4x3+2.

#### To determine

**To check:** The answer to part (a) is reasonable by comparing the graphs of f and f′.

#### Answer

The answer to part (a) is reasonable.

#### Explanation

**Graph:**

Use the online graphing calculator to draw the graph of f(x) and f′(x) as shown below in Figure 1.

**Observation:**

From Figure 1, it is observed that the function f(x)=x4+2x has a horizontal tangent at x=−123 and f′(x)=0 at x=−123. That is,

f′(x)=4(−123)3+2=4(−12)+2=−2+2=0

Thus, the derivative f′(x)=0 when *f* has a horizontal tangent.

Also, the value of f′(x)>0 when slope of f(x) has positive value and f′(x)<0 when slope of f(x) has negative value.

Thus, it can be concluded that the answer to part (a) is reasonable.