#### To determine

**To find:** The differentiation of the function y=t3+3tt2−4t+3.

#### Answer

The differentiation of the function y=t3+3tt2−4t+3 is. dydx=t4−8t3+6t2+9(t−3)2(t−1)2.

#### Explanation

**Given:**

The function y=t3+3tt2−4t+3.

**Derivative rule:**

(1) Quotient Rule: ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]−f1(x)ddx[f2(x)][f2(x)]2

(2) Constant multiple rule: ddx(c⋅f)=c⋅ddx(f)

(3) Power Rule: ddx(xn)=nxn−1

(4) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

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

**Calculation:**

The derivative of the function is dydx, which is obtained as follows.

dydt=ddt(y)=ddt(t3+3tt2−4t+3)

Use quotient rule (1) and simplify the terms,

dydx=(t2−4t+3)ddt[t3+3t]−(t3+3t)ddt[t2−4t+3][t2−4t+3]2

Apply the derivative rules (4), (5), (2), and (3) and simplify the terms,

dydt=(t2−4t+3)[ddt(t3)+ddt(3t)]−(t3+3t)[ddt(t2)−ddt(4t)+ddt(3)][t2−4t+3]2=(t2−4t+3)[ddt(t3)+3ddt(t)]−(t3+3t)[ddt(t2)−4ddt(t)+ddt(3)][t2−4t+3]2=(t2−4t+3)[3t2+3]−(t3+3t)[2t−4+0][t2−4t+3]2

Expand the numerator and simplify further,

dydt=(3t4−12t3+12t2−12t+9)−(2t4+6t2−4t3−12t)[t2−4t+3]2=(3t4−12t3+12t2−12t+9)+(−2t4+4t3−6t2+12t)[t2−4t+3]2=t4−8t3+6t2+9(t−3)2(t−1)2

Therefore, the differentiation of the function y=t3+3tt2−4t+3 is dydx=t4−8t3+6t2+9(t−3)2(t−1)2.