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.