Exam for Total Differentiation

The total differentiation of a function, f of several variables. For example, t, x, y etc., with respect to t is different from the partial derivative. The total differentiation of f with respect to t does not assume and other variables are constant while t varies. The  total differentiation adds in these indirect dependencies to find the overall dependency of f on t. For example, the total differentiation of f(t,x,y) with respect to t is

(df)/(dt) = (df)/(dt) +  ((df)/(dx)) ((dx)/(dt)) + ((df)/(dy))  ((dy)/(dt)) .

Exam for total differentiation problems:

Let us see some example problems for total differentiation and its helps to exam preparation.

Exam for total differentiation problem 1:

Find the total differentiation of trigonometric term (cot x2) with respect to x.

Solution:

Given trigonometric term is  (cot x2)

Let      y = (cot x2)

Putting x2 = u and  cot x2 = cot u = y

y = cot u,        and       u = x2

dy/(du) = - cosec2 u, and  (du)/(dx) = 2x

The total differentiation is  dy/dx

dy / dx = [  dy/(du) ×  (du)/(dx)]

= [( - cosec2 u) × 2x]

=  2x ( - cosec2 u)                                                         (u = x2)

= - 2x cosec2 (x2).

dy/dx = - 2x cosec2 (x2).

Answer:    dy/dx =  - 2x cosec2 (x2)

Exam for total differentiation problem 2:

Find the total differentiation of  function  P = xe^(y + z) ,  x = uv,  y = u - v , z = u + v

Find     (dP)/(dv)      when u = 1  and v = -1

Solution:

Given function is    P =  xe^(y+ z),

x = uv,        y = u - v        and     z = u + v

((dx)/(dv))= u = 1          ((dy)/(dv))  = -1     and    ((dz)/(du)) = 1

We know the chain rule of partial differentiation. it is the Total differentiation

 (dP)/(du) =  ((dP)/(dx)) ((dx)/(du)) +  ((dP)/(dy))   ((dy)/(du))  + ((dP)/(dz))   ((dz)/(du)) .

Substitute the u and v value we get   x = -1,    y = 2   and z = 0

P = xe^(y + z),

Differentiate P with respect to x .       (dP)/(dx)   = e^(y + z).

Differentiate P with respect to y .        (dP)/(dy) =  xe^(y + z)

Differentiate P with respect to z.       (dP)/(dz)   = xe^(y + z) .

Total differentiation is       (dP)/(dv) =  ((dP)/(dx)) ((dx)/(dv)) +  ((dP)/(dy))   ((dy)/(dv))  + ((dP)/(dz))   ((dz)/(dv)).

= e^(y + z).(1) +  xe^(y + z) .(-1) +  xe^(y + z).1

= e^(2 + 0).(1) +  (-1)e^(2 + 0) .(-1) +  (-1)e^(2 + 0).1

= e2 + e2 - e2

= e2

Answer:   e2