#### To determine

**To find:** The differentiation of the function y=1t3+2t2−1.

#### Answer

The differentiation of the function y=1t3+2t2−1 is dydt=−t(3t+4)(t3+2t2−1)2.

#### Explanation

**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:**

The derivative of the function y=1t3+2t2−1 is dydt, which is obtained as follow,

dydt=ddt(y)=ddt(1t3+2t2−1)

Apply the quotient rule (1),

dydt=[(t3+2t2−1)ddt(1)]−[(1)ddt(t3+2t2−1)](t3+2t2−1)2

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

dydt=[(t3+2t2−1)ddt(1)]−(1)[ddt(t3)+ddt(2t2)−ddt(1)](t3+2t2−1)2=[(t3+2t2−1)ddt(1)]−[ddt(t3)+2ddt(t2)−ddt(1)](t3+2t2−1)2

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

dydt=[(t3+2t2−1)(0)]−[(3t3−1)+(4t2−1)−0](t3+2t2−1)2=−3t2−4t(t3+2t2−1)2=−t(3t+4)(t3+2t2−1)2

Therefore, the differentiation of the function y=1t3+2t2−1 is dydt=−t(3t+4)(t3+2t2−1)2.