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) = e^{x + 3y} log (x² + y²), (-1, 1)

Solution:

(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) = e^{x + 3y} log (x² + y²), (-1, 1)

Question 2.

For each of the following functions find the f_{x}, and f_{y} and show that f_{xy} = f_{yx}

(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) = \(\frac { 3x }{ y+sinx }\)

(ii) f(x, y) = tan^{-1}(x/y)

(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)

∴ \(\frac{\partial^2 f}{\partial y \partial x}\) = \(\frac{\partial^2 f}{\partial x \partial y}\)

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:

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:

Question 5.

For each of the following functions find the g_{xy}, g_{xx}, g_{yy} and g_{yx}

(i) g(x, y) = x e^{y} + 3x²y

(ii) g(x, y) = log (5x + 3y)

(iii) g(x, y) = x² + 3xy – 7y + cos(5x)

Solution:

(i) g(x, y) = x e^{y} + 3x²y

(ii) g(x, y) = log (5x + 3y)

(iii) g(x, y) = x² + 3xy – 7y + cos(5x)

g_{x} = 2x+ 3y – sin 5x × 5

g_{x} = 2x+ 3y – 5 sin 5x

g_{xx} = 2 – 25 cos 5x

g_{yx} = 3

g_{y} = 3x – 7

g_{yy} = 0

g_{xy} = 3

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:

Question 7.

If V (x, y) = e^{x} ( x cosy – y siny), then Prove that \(\frac{\partial^2 V}{\partial x^2}\) + \(\frac{\partial^2 V}{\partial y^2}\) = 0

Solution:

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)

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

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