#### To determine

**To find:** The derivative of
f(x),
f′(x) and to compare their graphs.

#### Answer

The derivative of
f(x)=3x15−5x3+3 is
45x14−15x2.

#### Explanation

**Given:**

The function is
f(x)=3x15−5x3+3

**Derivative rules:**

(1) Derivative of constant function:
ddx(c)=0

(2) Power Rule:
ddx(xn)=nxn−1

(3) Sum Rule:
ddx(f+g)=ddx(f)+ddx(g)

(4) Difference Rule:
ddx(f−g)=ddx(f)−ddx(g)

(5) Constant Multiple Rule:
ddx(cf)=cddx(f)

**Calculation:**

The derivative of
f(x) is
f′(x), which is obtained as follows.

f′(x)=ddx(f(x)) =ddx(3x15−5x3+3)

Apply the sum rule (3) and the difference rule (4),

f′(x)=ddx(3x15−5x3+3) =ddx(3x15)−ddx(5x3)+ddx(3)

Apply the derivative rule (5) and derivative rule (1),

f′(x)=3ddx(x15)−5ddx(x3)+ddx(3) =3ddx(x15)−5ddx(x3)+0

Apply the power rule (2) and simplify the expression,

f′(x)=3(15x15−1)−5(3x3−1)=3(15x14)−5(3x2)=45x14−15x2

Therefore, the derivation of the function
f(x)=3x15−5x3+3 is
45x14−15x2.

To sketch the graphs of the functions
f(x) and
f′(x) the functions are
f(x)=3x15−5x3+3 and f′(x)= 45x14−15x2.

**Graph:**

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

**Observation:**

From Figure 1, it is noticed that, the function
f(x) is a decreasing function between the region
x=−1 and
x=1.

The function
f(x) is an increasing function outside
x=−1 and
x=1,
f′(x) is going straight upwards in the region
x=−1 and
x=1.

In the region
x=−1 and
x=1, the slope of
f(x) is negative which then becomes straight and then comes back to the negative slope whereas the slope of
f′(x) becomes negative and it reaches zero and then again has a negative slope.

When the graph approaches to distant right or left then
f′(x) tends to
∞ whereas the graph of
f(x) has a positive slop.

Therefore, it can be concluded that the derivative of the function is reasonable.