#### To determine

**To find:** The differentiation of the function
u=(1t−1t)2.

#### Answer

The differentiation of the function
u=(1t−1t)2 is
−2t3+3t2t−1t2.

#### Explanation

**Formula used:**

**The Constant Multiple Rule:**

If *c* is a constant and
f(v) is a differentiable function, then the constant multiple rule is,

ddv[c⋅f(v)]=c⋅ddvf(v) (1)

**The Power Rule:**

If *n* is any real number, then the power rule is,

ddv(vn)=nvn−1 (2)

**The Difference Rule:**

If
f(v) and
g(v) are both differentiable functions, then the difference rule is,

ddv[f(v)−g(v)]=ddv(f(v))−ddv(g(v)) (3)

**Calculation:**

The derivative of
u=(1t−1t)2 is
u′, which is obtained below:

u′=ddt(u) =ddt((1t−1t)2)=ddt((t−1−t−12)2)=ddt(t−2−2t−1+(−12)+t−1)

Apply the difference rule as shown in equation (3).

u′=ddt(t−2−2t−32+t−1)=ddt(t−2)−ddt(2t−32)+ddt(t−1)

Apply constant multiple rule as shown in equation (1),

ddt(2t−32)=2ddt(t−32)

Apply power rule as shown in equation (2),

u′=−2t−2−1−2(−32t−32−1)+(−1)t−1−1=−2t−3+3t−52−t−2

The variable of the equation
−2t−3+3t−52−t−2 can be replaced in the denominator as,

−2t−3+3t−52−t−2=−2t3+3t52−1t2=−2t3+3t12+42−1t2=−2t3+3t12t42−1t2=−2t3+3t2t−1t2

Therefore, the differentiation of the function
u=(1t−1t)2 is
−2t3+3t2t−1t2.