#### To determine

**To find:** The first and second derivatives of the function.

#### Answer

The first derivative of the function f(x) is 0.005x4−0.06x2_.

The second derivative of f(x) is 0.02x3−0.12x_.

#### Explanation

**Given:**

The function f(x)=0.001x5−0.02x3

**Derivative rules:**

(1) Constant Multiple Rule: ddx[c⋅f(x)]=c⋅ddxf(x)

(2) Difference Rule: ddx[f(x)−g(x)]=ddx(f(x))−ddx(g(x))

(3) Power Rule: ddx(xn)=nxn−1

**Calculation:**

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

f′(x)=ddx(f(x)) =ddx(0.001x5−0.02x3)

Apply the difference rule (2),

f′(x)=ddx(0.001x5)−ddx(0.02x3)

Apply the constant multiple rule (1),

f′(x)=0.001ddx(x5)−0.02ddx(x3)

Apply the power rule (3) and simplify the expressions.

f′(x)=0.001(5x5−1)−0.02(3x3−1)=0.001(5x4)−0.02(3x2)=0.005x4−0.06x2

Therefore, the first derivative of the function f(x) is 0.005x4−0.06x2_ .

The second derivative of f(x) is f″(x), which is obtained as follows,

f″(x)=ddx(f′(x))=ddx(0.005x4−0.06x2)

Apply the difference rule (2),

f″(x)=ddx(0.005x4)−ddx(0.06x2)

Apply the constant multiple rule (1),

f″(x)=0.005ddx(x4)−0.06ddx(x2)

Apply the power rule (3) and simplify the expressions.

f″(x)=0.005(4x4−1)−0.06(2x2−1)=0.005(4x3)−0.06(2x)=0.02x3−0.12x

Therefore, the second derivative of f(x) is 0.02x3−0.12x_.