Question Bank

Unit–3

Partial Differentiation

Question-1: Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50)cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xycos(y²x + 5x – 8)

Question-2: if , then show that-

Sol. Here we have,

u =     …………………..(1)

Now partially differentiate eq.(1) w.r to x and y , we get

                                               =

Or

                 ………………..(2)

And now,

                                                =

         ………………….(3)

Adding eq. (1) and (3) , we get

=  0

Hence proved.

Question-3: If u = x²(y-x) + y²(x-y), then show that  -2 (x – y)²

Solution - here,  u = x²(y-x) + y²(x-y)

                                u = x²y - x³ + xy² - y³,

Now differentiate u partially with respect to x and y respectively,

=  2xy – 3x² + y²        --------- (1)

= x² + 2xy – 3y²          ---------- (2)

Now adding equation (1) and (2), we get

= -2x² - 2y² + 4xy

                                               = -2 (x² + y² - 2xy)

                                               = -2 (x – y)²

Question-4: Find      if u = x³y⁴   where   x = t³     and    y = t².

Sol. As we know that by definition,    =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Example-3: if w = x² + y – z + sintand  x + y = t, find

                   (a)    y,z

(b)   t, z

Sol. With x, y, z independent, we have

t = x + y, w = x²+ y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

=  2x + cos (x + y)

With x, t, z independent, we have

                                                          Y = t-x,   w=  x² + (t-x) + sin t

Thust, z   =  2x - 1

Question-5: If z is the function of x and y , and x =   , y = , then prove that,

Sol. Here , it is given that,  z is the function of x and y & x , y are the functions of u and v.

So that,

  ……………….(1)

And,

   ………………..(2)

Also there is,

      x =   and  y = ,

Now,

   ,       ,      ,    

From equation(1) , we get

  ……………….(3)

And from eq. (2) , we get

    …………..(4)

Subtracting eq. (4) from (3), we get

= ) – (

              = x

Hence proved.

Question-6: If where then find the value of ?

Sol.

Given  

Where 

By chain rule

                          Now substituting the value of x ,y,z we get

-6

8

Question-7:If where the relation is .

Find the value of

Let the given relation is denoted by

We know that

        Differentiating u with respect to x and using chain rule

Question-8: If x = r sin , y = r sin , z = r cos, then show that

sin  also find 

Sol. 

We know that,

                                       =

                                       =

                                       = ( on solving the determinant)

                                      = 

Now using first propert of Jacobians, we get

Question-9: Find the Taylor’s expansion of about (1 , 1) up to second degree term.

Sol. We have,

At (1 , 1)

Now by using Taylor’s theorem-

……

Suppose 1 + h = x  then h =  x – 1

                 1 + k = y   then k = y - 1

……

  =

……..

Question-10: Find out the maxima and minima of the function

Sol.

Given …(i)

         Partially differentiating (i) with respect to x we get

….(ii)

         Partially differentiating (i) with respect to y we get

….(iii)

          Now, form the equations

                    Using (ii) and (iii) we get

using above two equations

                     Squaring both side we get

                      Or 

                       This show that

                         Also we get

                 Thus we get the pair of value as

                  Now, we calculate

      Putting above values in

      At point (0,0)  we get

      So, the point (0,0) is a saddle point.

      At point we get

        So the point is the minimum point where

          In case 

         So the point is the maximum point where