Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 1.
Find the partial dervatives of the following functions at indicated points.
(i) f(x, y) = 3x² – 2xy + y² + 5x + 2, (2, -5)
(ii) g(x, y) = 3x² + y² + 5x + 2, (2, -5)
(iii) h(x, y, z) = x sin (xy) + z² x, (2, π/4, 1)
(iv) G(x, y) = ex + 3y log (x² + y²), (-1, 1)
Solution:
(i) f(x, y) = 3x² – 2xy + y² + 5x + 2, (2, -5)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 1

(ii) g(x, y) = 3x² + y² + 5x + 2, (2, -5)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 2

(iii) h(x, y, z) = x sin (xy) + z² x, (2, π/4, 1)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 3

(iv) G(x, y) = ex + 3y log (x² + y²), (-1, 1)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 4

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 2.
For each of the following functions find the fx, and fy and show that fxy = fyx
(i) f(x, y) = \(\frac { 3x }{ y+sinx }\)
(ii) f(x, y) = tan-1(x/y)
(iii) f(x, y) = cos (x² – 3xy)
Solution:
(i) f(x, y) = \(\frac { 3x }{ y+sinx }\)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 5

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 6

(ii) f(x, y) = tan-1(x/y)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 7

(iii) f(x, y) = cos (x² – 3xy)
\(\frac{\partial f}{\partial x}\) = -sin (x² – 3xy) × (2x – 3y)
\(\frac{\partial^2 f}{\partial y \partial x}\) = – sin (x² – 3xy) x – 3 + (2x – 3y) × [-cos (x² – 3xy)] × – 3x
= 3 sin (x² – 3xy) + 3x (2x – 3y) cos (x² – 3xy)
\(\frac{\partial f}{\partial x}\) = – sin (x² – 3xy) × -3x
= 3x sin (x² – 3xy)
\(\frac{\partial^2 f}{\partial x \partial y}\) = 3sin (x² – 3xy) + 3x cos (x² – 3xy) (2x – 3y)
∴ \(\frac{\partial^2 f}{\partial y \partial x}\) = \(\frac{\partial^2 f}{\partial x \partial y}\)

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 3.
If U (x, y, z) = \(\frac { x^2+y^2 }{ xy }\) + 3z²y, find \(\frac{\partial U}{\partial x}\), \(\frac{\partial U}{\partial y}\) and \(\frac{\partial U}{\partial z}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 8

Question 4.
If U(x, y, z) = log (x³ + y³ + z³) find
\(\frac{\partial U}{\partial x}\) + \(\frac{\partial U}{\partial y}\) + \(\frac{\partial U}{\partial z}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 9

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 5.
For each of the following functions find the gxy, gxx, gyy and gyx
(i) g(x, y) = x ey + 3x²y
(ii) g(x, y) = log (5x + 3y)
(iii) g(x, y) = x² + 3xy – 7y + cos(5x)
Solution:
(i) g(x, y) = x ey + 3x²y
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 10

(ii) g(x, y) = log (5x + 3y)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 11

(iii) g(x, y) = x² + 3xy – 7y + cos(5x)
gx = 2x+ 3y – sin 5x × 5
gx = 2x+ 3y – 5 sin 5x
gxx = 2 – 25 cos 5x
gyx = 3
gy = 3x – 7
gyy = 0
gxy = 3

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 6.
Let w(x, y, z) = \(\frac { 1 }{ \sqrt{x^2+y^2+z^2} }\) = 1, (x, y, z) ≠ (0, 0, 0), show that \(\frac{\partial^ w}{\partial x^2}\) + \(\frac{\partial^2 w}{\partial y^2}\) + \(\frac{\partial^2 w}{\partial z^2}\) = 0
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 12
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 13

Question 7.
If V (x, y) = ex ( x cosy – y siny), then Prove that \(\frac{\partial^2 V}{\partial x^2}\) + \(\frac{\partial^2 V}{\partial y^2}\) = 0
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 14

Question 8.
If w (x, y) = xy + sin (xy), then Prove that \(\frac{\partial^2 w}{\partial y \partial x}\) = \(\frac{\partial^2 w}{\partial x \partial y}\)
Solution:
Now, w (x, y) = xy + sin (xy)
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 15

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

Question 9.
If v(x, y, z) = x³ + y³ + z³ +3xyz, Show that \(\frac{\partial^2 v}{\partial y \partial z}\) = \(\frac{\partial^2 v}{\partial z \partial y}\)
Solution:
v (x, y, z) = x³ + y³ + z³ + 3xyz
Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4 16

Question 10.
A from produces two types of calculates each week, x number of type A and y number of type B. The weekly revenue and cost functions = (in rupees) are
R (x, y) = 80x + 90y + 0.04xy – 0.05x² – 0.05y² and C (x, y) = 8x + 6y + 2000 respectively.
(i) Find the Profit function P(x, y).
(ii) Find \(\frac{\partial P}{\partial x}\) (1200, 1800) and \(\frac{\partial P}{\partial y}\) (1200, 1800) and interpret these results.
Solution:
(i) P (x, y) = R (x, y) – C (x, y)
= (80x + 90y + 0.04xy – 0.05x² – 0.05y²) – (8x + 6y + 2000)
P (x, y) = 72x + 84y + 0.04xy – 0.05x² – 0.05y² – 2000

(ii) \(\frac{\partial P}{\partial x}\) = 72 + 0.04y – 0.1 x
\(\frac{\partial P}{\partial x}\) (1200, 1800) = 72 + 0.04 × 1800 – 0.1 × 1200
= 72 + 72 – 120
= 144 – 120
= 24
\(\frac{\partial P}{\partial y}\) = 84 + 0.04x – 0.1 y
\(\frac{\partial P}{\partial y}\) (1200, 1800) = 84 + 0.04 × 1200 – 0.1 × 1800
= 84 + 48 – 180
= 132 – 180
= -48

Samacheer Kalvi 12th Maths Guide Chapter 8 Differentials and Partial Derivatives Ex 8.4

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