#### To determine

**To find:** The *n*th derivative of the function.

#### Answer

The *n*th derivative of the function f(x)=xn is n(n−1)(n−2)...2⋅1=n!_.

#### Explanation

**Given:**

The function is f(x)=xn

**Derivative rules:**

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn−1

**Calculation:**

The first derivative of f(x)=xn is

f′(x)=ddx(xn)

By the power rule (2), f′(x)=nxn−1.

Thus, the first derivative of f(x)=xn is f′(x)=nxn−1_.

The second derivative of f(x)=xn is,

f″(x)=ddx(f′(x))=ddx(nxn−1)

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

f″(x)=nddx(xn−1)=n((n−1)x(n−1)−1)=n(n−1)xn−1−1=n(n−1)xn−2

Thus, the second derivative of f(x)=xn is n(n−1)xn−2_.

The third derivative of f(x)=xn is,

f‴(x)=ddx(f″(x))=ddx(n(n−1)xn−2)

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

f‴(x)=n(n−1)ddx(xn−2)=n(n−1)((n−2)x(n−2)−1)=n(n−1)(n−2)xn−2−1=n(n−1)(n−2)xn−3

Thus, the third derivative of f(x)=xn is n(n−1)(n−2)xn−3_.

From the first, second and the third derivatives of f(x)=xn, the general *k*th form of f(x)=xn can be expressed as follows,

fk(x)=n(n−1)(n−2)...(n−k+1)xn−k

Substitute *n* for *k* in the above equation, then the *n*th derivative of f(x)=xn is

fn(x)=n(n−1)(n−2)...2⋅1⋅xn−n=n(n−1)(n−2)...2⋅1⋅x0=n(n−1)(n−2)...2⋅1=n!

Therefore, it can be concluded that the *n*th derivative of f(x)=xn is n(n−1)(n−2)...2⋅1=n!_.

#### To determine

**To find:** The *n*th derivative of the function.

#### Answer

The *n*th derivative of f(x)=1x is. (−1)nn!xn+1_.

#### Explanation

**Given:**

The function is f(x)=1x

**Derivative rules:**

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn−1

**Calculation:**

The first derivative of f(x)=1x is obtained as follows.

f′(x)=ddx(1x)=ddx(x−1)

Apply the power rule (2),

f′(x)=(−1)x−1−1=−x−2

Thus, the first derivative of f(x)=1x is f′(x)=−x−2.

The second derivative of f(x)=1x is obtained as follows.

f″(x)=ddx(f′(x))=ddx(−x−2)

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

f″(x)=(−1)ddx(x−2)=−1((−2)x−2−1)=2x−3

Thus, the second derivative of f(x)=1x is f″(x)=2x−3.

The third derivative of f(x)=1x is obtained as follows.

f‴(x)=ddx(f″(x))=ddx(2x−3)

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

f′′′(x)=2ddx(x−3)=2((−3)x−3−1)=−6x−4

Thus, the third derivative of f(x)=1x is f′′′(x)=−6x−4.

The first, second and third derivatives of f(x)=1x can be expressed as follows,

f′(x)=(−1)1!x−2f″(x)=(−1)22!x−3f‴(x)=(−1)33!x−4

Proceed in the similar way, the *n*th derivative of f(x)=1x is f(n)(x)=(−1)nn!xn+1.

Therefore, it can be concluded that the *n*th derivative of f(x)=1x is f(n)(x)=(−1)nn!xn+1.