Question Bank (unit-5)
Question-1: - Calculate
and for the following functionf(x , y) = 3x³-5y²+2xy-8x+4y-20
Sol. To calculate treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,
= [3x³-5y²+2xy-8x+4y-20]
= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]
= 9x² - 0 + 2y – 8 + 0 – 0
= 9x² + 2y – 8
Similarly partial derivative of f(x,y) with respect to y is:
= [3x³-5y²+2xy-8x+4y-20]
= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]
= 0 – 10y + 2x – 0 + 4 – 0
= 2x – 10y +4.
Question-2: 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)
= 2xy cos(y²x + 5x – 8)
Question-3 : 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-4: Example-3: if w = x² + y – z + sint and 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
thus t, z = 2x – 1
Question-5: if φ( cx – az , cy – bz) = 0 then show that ap + bq = c
Where p = q =
Sol. We have,
φ( cx – az , cy – bz) = 0
φ( r , s) = 0
Where,
We know that,
Again we do,
By adding the two results, we get
Question-6: : 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-7: 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-8: If u = xy + sin(xy), show that = .
Solution – u = xy + sin(xy)
= y+ ycos(xy)
= x+ xcos(xy)
x (- sin(xy).(y)) + cos(xy)
= 1 – xysin(xy) + cos(xy) -------------- (1)
1 + cos(xy) + y(-sin(xy) x)
= 1 – xy sin(xy) + cos(xy) -----------------(2)
From equation (1) and (2),
=
Question-9: If u(x,y,z) = log( tan x + tan y + tan z) , then prove that ,
Sol. Here we have,
u(x,y,z) = log( tan x + tan y + tan z) ………………..(1)
Diff. Eq.(1) w.r.t. x , partially , we get
……………..(2)
Diff. Eq.(1) w.r.t. y , partially , we get
………………(3)
Diff. Eq.(1) w.r.t. z , partially , we get
……………………(4)
Now multiply eq. 2 , 3 , 4 by sin 2x , sin 2y , sin 2z respectively and adding , in order to get the final result,
We get,
=
So that,
hence proved.
Question-10: If u = u(), then prove that .
Sol. We are given that,
u = u () = u( r , s)
Where r = and s =
Or we can write as-
r = and s =
Differentiate them partially with respect to x , y and z. , we get
As we know that-
Then,
Adding all these results, we get
.