#### To determine

**To find:** The derivative of the function F(x)=x4−5x3+xx2.

#### Answer

The derivative of the function F(x)=x4−5x3+xx2 is. 2x−5−32x−52.

#### Explanation

**Given:**

The function F(x)=x4−5x3+xx2.

**Derivative rule:**

(1) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2(x)]2

(2) Power Rule: ddx(xn)=nxn−1

(3). Sum rule: ddx(f+g)=ddx(f)+ddx(g)

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

(5) Difference rule: ddx(f−g)=ddx(f)−ddx(g)

**Calculation:**

**Method 1:**

Obtain the derivative of F(x) by using Quotient Rule.

The derivative of the function F(x) is F′(x), which is obtained as follows.

F′(x)=ddx(F(x))=ddx(x4−5x3+xx2)

Substitute x4−5x3+x for f1(x) and x2 for f2(x) in the quotient rule (1),

F′(x)=(x2)ddx(x4−5x3+x)−(x4−5x3+x)ddx(x2)(x2)2

Apply the derivative rules (3), (4), and (5),

F′(x)=(x2)[ddx(x4)−ddx(5x3)+ddx(x12)]−[(x4−5x3+x)ddx(x2)](x2)2=x2[ddx(x4)−5ddx(x3)+ddx(x12)]−[(x4−5x3+x12)ddx(x2)]x4

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

F′(x)=x2[4x3−5(3x3−1)+(12x12−1)]−(x4−5x3+x12)2xx4=x2[4x3−15x2+12x−12]−(x4−5x3+x12)2xx4=4x5−15x4+12x2−12−2x5+10x4−2x1+12x4=4x5−15x4+12x32−2x5+10x4−2x32x4

Simplify the numerator and obtain the differentiation of F(x).

F′(x)=2x5−5x4−32x32x4=2x−5−32x32−4=2x−5−32x−52

Therefore, the differentiation of the function F(x)=x4−5x3+xx2 is 2x−5−32x−52.

**Method 2:**

Obtain the derivative of F(x) by simplifying first.

Simplify the function F(x),

F(x)=x4−5x3+x12x2=x4x2−5x3x2+x12x2=x2−5x+x12−2=x2−5x+x−32

The derivative of the function F(x) is F′(x), which is obtained as follows.

F′(x)=ddx(F(x))=ddx(x2−5x+x−32)

Apply the derivative rules (3), (4) and (5),

F′(x)=ddx(x2)−ddx(5x)+ddx(x−32)=ddx(x2)−5ddx(x)+ddx(x−32)

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

F′(x)=2x−5(1)−32x−32−1=2x−5−32x−52

Thus, the differentiation of the function F(x)=x4−5x3+xx2 is 2x−5−32x−52.

Therefore, it can be concluded that the derivative of F(x) by both the methods **1** and **2** are same and method **2** seems to be the better one as it has lesser calculations.