Samacheer Kalvi Class 12 Maths Solutions Chapter 8 Differentials and Partial Derivatives Exercise 8.4

Question 1. Find the partial derivatives of the following functions at indicated points.
(i) f(x, y) = \( 3x^2 – 2xy + y^2 + 5x + 2, (2, -5) \)
(ii) g(x, y) = \( 3x^2 + y^2 + 5x + 2, (2, -5) \)
(iii) h(x, y, z) = \( x \sin (xy) + z^2 x, (2, \pi/4, 1) \)
(iv) G(x, y) = \( e^x + 3y \log (x^2 + y^2), (-1, 1) \)
Answer:
(i) For \( f(x, y) = 3x^2 – 2xy + y^2 + 5x + 2 \):
First, we find the partial derivative with respect to x:
\( \frac{\partial f}{\partial x} = 6x - 2y + 5 \)
Now, we evaluate this at the point (2, -5):
\( \frac{\partial f}{\partial x} (2, -5) = 6(2) - 2(-5) + 5 \)
\( = 12 + 10 + 5 = 27 \)
Next, we find the partial derivative with respect to y:
\( \frac{\partial f}{\partial y} = -2x + 2y \)
Evaluate this at the point (2, -5):
\( \frac{\partial f}{\partial y} (2, -5) = -2(2) + 2(-5) \)
\( = -4 - 10 = -14 \)
Partial derivatives measure how a function changes as one variable changes, while others are held constant.
(ii) For \( g(x, y) = 3x^2 + y^2 + 5x + 2 \):
First, we find the partial derivative with respect to x:
\( \frac{\partial g}{\partial x} = 6x + 5 \)
Now, we evaluate this at the point (2, -5) – *note: the source shows evaluation at (1,-2) for g. We will follow the question's stated point (2,-5).*: We use the given point (2, -5) to maintain consistency with the problem statement, even if the intermediate steps in the source document might show different values. Substituting (2, -5):
\( \frac{\partial g}{\partial x} (2, -5) = 6(2) + 5 \)
\( = 12 + 5 = 17 \)
Next, we find the partial derivative with respect to y:
\( \frac{\partial g}{\partial y} = 2y \)
Evaluate this at the point (2, -5):
\( \frac{\partial g}{\partial y} (2, -5) = 2(-5) \)
\( = -10 \)
(iii) For \( h(x, y, z) = x \sin (xy) + z^2 x \):
First, we find the partial derivative with respect to x (using the product rule and chain rule):
\( \frac{\partial h}{\partial x} = (1)\sin(xy) + x(\cos(xy) \cdot y) + z^2(1) \)
\( = \sin(xy) + xy \cos(xy) + z^2 \)
Now, we evaluate this at the point \( (2, \pi/4, 1) \):
\( \frac{\partial h}{\partial x} (2, \pi/4, 1) = \sin(2 \cdot \pi/4) + (2 \cdot \pi/4) \cos(2 \cdot \pi/4) + (1)^2 \)
\( = \sin(\pi/2) + (\pi/2) \cos(\pi/2) + 1 \)
\( = 1 + (\pi/2) \cdot 0 + 1 \)
\( = 1 + 0 + 1 = 2 \)
Next, we find the partial derivative with respect to y (treating x and z as constants):
\( \frac{\partial h}{\partial y} = x(\cos(xy) \cdot x) \)
\( = x^2 \cos(xy) \)
Evaluate this at the point \( (2, \pi/4, 1) \):
\( \frac{\partial h}{\partial y} (2, \pi/4, 1) = (2)^2 \cos(2 \cdot \pi/4) \)
\( = 4 \cos(\pi/2) \)
\( = 4 \cdot 0 = 0 \)
Finally, we find the partial derivative with respect to z (treating x and y as constants):
\( \frac{\partial h}{\partial z} = x(0) + (2z)(x) \)
\( = 2zx \)
Evaluate this at the point \( (2, \pi/4, 1) \):
\( \frac{\partial h}{\partial z} (2, \pi/4, 1) = 2(1)(2) \)
\( = 4 \)
(iv) For \( G(x, y) = e^x + 3y \log (x^2 + y^2) \):
First, we find the partial derivative with respect to x:
\( \frac{\partial G}{\partial x} = e^x + 3y \cdot \frac{1}{x^2+y^2} \cdot (2x) \)
\( = e^x + \frac{6xy}{x^2+y^2} \)
Now, we evaluate this at the point (-1, 1):
\( \frac{\partial G}{\partial x} (-1, 1) = e^{-1} + \frac{6(-1)(1)}{(-1)^2+(1)^2} \)
\( = e^{-1} + \frac{-6}{1+1} \)
\( = e^{-1} + \frac{-6}{2} \)
\( = e^{-1} - 3 \)
Next, we find the partial derivative with respect to y:
\( \frac{\partial G}{\partial y} = 0 + (3)\log(x^2+y^2) + 3y \cdot \frac{1}{x^2+y^2} \cdot (2y) \)
\( = 3\log(x^2+y^2) + \frac{6y^2}{x^2+y^2} \)
Evaluate this at the point (-1, 1):
\( \frac{\partial G}{\partial y} (-1, 1) = 3\log((-1)^2+(1)^2) + \frac{6(1)^2}{(-1)^2+(1)^2} \)
\( = 3\log(1+1) + \frac{6}{1+1} \)
\( = 3\log(2) + \frac{6}{2} \)
\( = 3\log(2) + 3 \)
\( = 3(\log(2) + 1) \)
In simple words: To find a partial derivative, we treat all other variables as if they are constant numbers and only differentiate with respect to the chosen variable. Then, we put in the given numbers for x, y, and z to get a final value. These values tell us how sensitive the function is to changes in each variable at that specific point.

🎯 Exam Tip: Remember to use the product rule or chain rule carefully when differentiating terms that involve multiple variables. Pay close attention to the point at which the derivative needs to be evaluated.

Question 2. For each of the following functions find the \( f_x \), and \( f_y \) and show that \( f_{xy} = f_{yx} \).
(i) \( f(x, y) = \frac { 3x }{ y+\sin x } \)
(ii) \( f(x, y) = \tan^{-1}(x/y) \)
(iii) \( f(x, y) = \cos (x^2 – 3xy) \)
Answer:
(i) For \( f(x, y) = \frac { 3x }{ y+\sin x } \):
First, find \( f_x \), the partial derivative with respect to x:
\( f_x = \frac{\partial}{\partial x} \left( \frac{3x}{y+\sin x} \right) \)
Using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \):
\( u = 3x \implies u' = 3 \)
\( v = y+\sin x \implies v' = \cos x \)
\( f_x = \frac{3(y+\sin x) - 3x(\cos x)}{(y+\sin x)^2} \)
\( = \frac{3y+3\sin x-3x\cos x}{(y+\sin x)^2} \)
Next, find \( f_y \), the partial derivative with respect to y:
\( f_y = \frac{\partial}{\partial y} \left( \frac{3x}{y+\sin x} \right) \)
Treat \( 3x \) as a constant:
\( f_y = 3x \frac{\partial}{\partial y} (y+\sin x)^{-1} \)
\( = 3x (-1)(y+\sin x)^{-2} (1) \)
\( = \frac{-3x}{(y+\sin x)^2} \)
Now, we find \( f_{xy} = \frac{\partial}{\partial y} (f_x) \):
\( f_{xy} = \frac{\partial}{\partial y} \left( \frac{3y+3\sin x-3x\cos x}{(y+\sin x)^2} \right) \)
Using the quotient rule again, with \( u = 3y+3\sin x-3x\cos x \implies u' = 3 \) (since \( \sin x \) and \( \cos x \) are constants with respect to y).
\( v = (y+\sin x)^2 \implies v' = 2(y+\sin x)(1) \)
\( f_{xy} = \frac{3(y+\sin x)^2 - (3y+3\sin x-3x\cos x) \cdot 2(y+\sin x)}{(y+\sin x)^4} \)
Factor out \( (y+\sin x) \) from the numerator:
\( f_{xy} = \frac{(y+\sin x)[3(y+\sin x) - 2(3y+3\sin x-3x\cos x)]}{(y+\sin x)^4} \)
\( = \frac{3y+3\sin x - 6y - 6\sin x + 6x\cos x}{(y+\sin x)^3} \)
\( = \frac{-3y-3\sin x+6x\cos x}{(y+\sin x)^3} \)
Finally, find \( f_{yx} = \frac{\partial}{\partial x} (f_y) \):
\( f_{yx} = \frac{\partial}{\partial x} \left( \frac{-3x}{(y+\sin x)^2} \right) \)
Using the quotient rule, with \( u = -3x \implies u' = -3 \)
\( v = (y+\sin x)^2 \implies v' = 2(y+\sin x)(\cos x) \)
\( f_{yx} = \frac{-3(y+\sin x)^2 - (-3x) \cdot 2(y+\sin x)(\cos x)}{(y+\sin x)^4} \)
Factor out \( (y+\sin x) \) from the numerator:
\( f_{yx} = \frac{(y+\sin x)[-3(y+\sin x) + 6x\cos x]}{(y+\sin x)^4} \)
\( = \frac{-3y-3\sin x+6x\cos x}{(y+\sin x)^3} \)
Thus, \( f_{xy} = f_{yx} \) is shown.

(ii) For \( f(x, y) = \tan^{-1}(x/y) \):
First, find \( f_x \), the partial derivative with respect to x:
\( f_x = \frac{\partial}{\partial x} (\tan^{-1}(x/y)) \)
Using the chain rule \( \frac{d}{du} \tan^{-1}(u) = \frac{1}{1+u^2} \frac{du}{dx} \), where \( u = x/y \).
\( f_x = \frac{1}{1+(x/y)^2} \cdot \frac{\partial}{\partial x} (x/y) \)
\( = \frac{1}{1+x^2/y^2} \cdot \frac{1}{y} \)
\( = \frac{y^2}{y^2+x^2} \cdot \frac{1}{y} \)
\( = \frac{y}{x^2+y^2} \)
Next, find \( f_y \), the partial derivative with respect to y:
\( f_y = \frac{\partial}{\partial y} (\tan^{-1}(x/y)) \)
\( f_y = \frac{1}{1+(x/y)^2} \cdot \frac{\partial}{\partial y} (x/y) \)
\( = \frac{1}{1+x^2/y^2} \cdot x(-1)y^{-2} \)
\( = \frac{y^2}{y^2+x^2} \cdot \frac{-x}{y^2} \)
\( = \frac{-x}{x^2+y^2} \)
Now, find \( f_{xy} = \frac{\partial}{\partial y} (f_x) \):
\( f_{xy} = \frac{\partial}{\partial y} \left( \frac{y}{x^2+y^2} \right) \)
Using the quotient rule, with \( u = y \implies u' = 1 \)
\( v = x^2+y^2 \implies v' = 2y \)
\( f_{xy} = \frac{1(x^2+y^2) - y(2y)}{(x^2+y^2)^2} \)
\( = \frac{x^2+y^2 - 2y^2}{(x^2+y^2)^2} \)
\( = \frac{x^2-y^2}{(x^2+y^2)^2} \)
Finally, find \( f_{yx} = \frac{\partial}{\partial x} (f_y) \):
\( f_{yx} = \frac{\partial}{\partial x} \left( \frac{-x}{x^2+y^2} \right) \)
Using the quotient rule, with \( u = -x \implies u' = -1 \)
\( v = x^2+y^2 \implies v' = 2x \)
\( f_{yx} = \frac{-1(x^2+y^2) - (-x)(2x)}{(x^2+y^2)^2} \)
\( = \frac{-x^2-y^2 + 2x^2}{(x^2+y^2)^2} \)
\( = \frac{x^2-y^2}{(x^2+y^2)^2} \)
Thus, \( f_{xy} = f_{yx} \) is shown.

(iii) For \( f(x, y) = \cos (x^2 – 3xy) \):
First, find \( f_x \), the partial derivative with respect to x:
\( f_x = \frac{\partial}{\partial x} (\cos (x^2 – 3xy)) \)
Using the chain rule:
\( f_x = -\sin(x^2-3xy) \cdot \frac{\partial}{\partial x}(x^2-3xy) \)
\( = -\sin(x^2-3xy) \cdot (2x-3y) \)
Next, find \( f_y \), the partial derivative with respect to y:
\( f_y = \frac{\partial}{\partial y} (\cos (x^2 – 3xy)) \)
Using the chain rule:
\( f_y = -\sin(x^2-3xy) \cdot \frac{\partial}{\partial y}(x^2-3xy) \)
\( = -\sin(x^2-3xy) \cdot (-3x) \)
\( = 3x \sin(x^2-3xy) \)
Now, find \( f_{xy} = \frac{\partial}{\partial y} (f_x) \):
\( f_{xy} = \frac{\partial}{\partial y} (-\sin(x^2-3xy) \cdot (2x-3y)) \)
Using the product rule \( (uv)' = u'v + uv' \):
\( u = -\sin(x^2-3xy) \implies u' = -\cos(x^2-3xy) \cdot (-3x) = 3x\cos(x^2-3xy) \)
\( v = (2x-3y) \implies v' = -3 \)
\( f_{xy} = (3x\cos(x^2-3xy))(2x-3y) + (-\sin(x^2-3xy))(-3) \)
\( = 3x(2x-3y)\cos(x^2-3xy) + 3\sin(x^2-3xy) \)
Finally, find \( f_{yx} = \frac{\partial}{\partial x} (f_y) \):
\( f_{yx} = \frac{\partial}{\partial x} (3x \sin(x^2-3xy)) \)
Using the product rule \( (uv)' = u'v + uv' \):
\( u = 3x \implies u' = 3 \)
\( v = \sin(x^2-3xy) \implies v' = \cos(x^2-3xy) \cdot (2x-3y) \)
\( f_{yx} = 3\sin(x^2-3xy) + 3x\cos(x^2-3xy)(2x-3y) \)
\( = 3\sin(x^2-3xy) + 3x(2x-3y)\cos(x^2-3xy) \)
Thus, \( f_{xy} = f_{yx} \) is shown. This property of mixed partial derivatives being equal is known as Clairaut's theorem, and it holds true for most well-behaved functions found in calculus.
In simple words: For each function, we first find how it changes when only 'x' changes (this is \( f_x \)), and how it changes when only 'y' changes (this is \( f_y \)). Then we find \( f_{xy} \) which means taking the 'x' derivative and then the 'y' derivative. We also find \( f_{yx} \) which means taking the 'y' derivative and then the 'x' derivative. In all these cases, the results are the same, showing a common rule in calculus where the order of taking these changes does not matter for most smooth functions.

🎯 Exam Tip: When showing \( f_{xy} = f_{yx} \), compute each mixed partial derivative step-by-step. Remember that Clairaut's Theorem guarantees equality if the second partial derivatives are continuous in the region of interest.

Question 3. If \( U(x, y, z) = \frac{x^2+y^2}{xy} + 3z^2y \), find \( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \) and \( \frac{\partial U}{\partial z} \).
Answer:
Given the function \( U(x, y, z) = \frac{x^2+y^2}{xy} + 3z^2y \).
First, we can simplify the first term: \( \frac{x^2+y^2}{xy} = \frac{x^2}{xy} + \frac{y^2}{xy} = \frac{x}{y} + \frac{y}{x} = xy^{-1} + yx^{-1} \).
So, \( U(x, y, z) = xy^{-1} + yx^{-1} + 3z^2y \).

Now, find \( \frac{\partial U}{\partial x} \):
Treat y and z as constants.
\( \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(xy^{-1}) + \frac{\partial}{\partial x}(yx^{-1}) + \frac{\partial}{\partial x}(3z^2y) \)
\( = (1)y^{-1} + y(-1)x^{-2} + 0 \)
\( = y^{-1} - yx^{-2} \)
\( = \frac{1}{y} - \frac{y}{x^2} \)

Next, find \( \frac{\partial U}{\partial y} \):
Treat x and z as constants.
\( \frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(xy^{-1}) + \frac{\partial}{\partial y}(yx^{-1}) + \frac{\partial}{\partial y}(3z^2y) \)
\( = x(-1)y^{-2} + (1)x^{-1} + 3z^2(1) \)
\( = -xy^{-2} + x^{-1} + 3z^2 \)
\( = \frac{-x}{y^2} + \frac{1}{x} + 3z^2 \)

Finally, find \( \frac{\partial U}{\partial z} \):
Treat x and y as constants.
\( \frac{\partial U}{\partial z} = \frac{\partial}{\partial z}(xy^{-1}) + \frac{\partial}{\partial z}(yx^{-1}) + \frac{\partial}{\partial z}(3z^2y) \)
\( = 0 + 0 + 3(2z)y \)
\( = 6zy \)
Partial derivatives are very useful for understanding how multi-variable functions change along different axes.
In simple words: We are looking at a mathematical rule for U which depends on x, y, and z. We find three new rules: one shows how U changes if only x moves, another shows how U changes if only y moves, and the third shows how U changes if only z moves. We do this by treating the other two letters as fixed numbers while working on one.

🎯 Exam Tip: Simplify the function first if possible, like \( \frac{x^2+y^2}{xy} \) to \( \frac{x}{y} + \frac{y}{x} \), as this often makes differentiation easier. Remember that when differentiating with respect to one variable, all other variables are treated as constants.

Question 4. If \( U(x, y, z) = \log (x^3 + y^3 + z^3) \), find \( \frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} + \frac{\partial U}{\partial z} \).
Answer:
Given the function \( U(x, y, z) = \log (x^3 + y^3 + z^3) \).

First, find \( \frac{\partial U}{\partial x} \):
Using the chain rule \( \frac{d}{dx} \log(f(x)) = \frac{f'(x)}{f(x)} \).
\( \frac{\partial U}{\partial x} = \frac{1}{x^3 + y^3 + z^3} \cdot \frac{\partial}{\partial x}(x^3 + y^3 + z^3) \)
\( = \frac{1}{x^3 + y^3 + z^3} \cdot (3x^2 + 0 + 0) \)
\( = \frac{3x^2}{x^3 + y^3 + z^3} \)

Next, find \( \frac{\partial U}{\partial y} \):
Similarly, treating x and z as constants.
\( \frac{\partial U}{\partial y} = \frac{1}{x^3 + y^3 + z^3} \cdot \frac{\partial}{\partial y}(x^3 + y^3 + z^3) \)
\( = \frac{1}{x^3 + y^3 + z^3} \cdot (0 + 3y^2 + 0) \)
\( = \frac{3y^2}{x^3 + y^3 + z^3} \)

Finally, find \( \frac{\partial U}{\partial z} \):
Similarly, treating x and y as constants.
\( \frac{\partial U}{\partial z} = \frac{1}{x^3 + y^3 + z^3} \cdot \frac{\partial}{\partial z}(x^3 + y^3 + z^3) \)
\( = \frac{1}{x^3 + y^3 + z^3} \cdot (0 + 0 + 3z^2) \)
\( = \frac{3z^2}{x^3 + y^3 + z^3} \)

Now, sum these partial derivatives:
\( \frac{\partial U}{\partial x} + \frac{\partial U}{\partial y} + \frac{\partial U}{\partial z} = \frac{3x^2}{x^3 + y^3 + z^3} + \frac{3y^2}{x^3 + y^3 + z^3} + \frac{3z^2}{x^3 + y^3 + z^3} \)
\( = \frac{3x^2 + 3y^2 + 3z^2}{x^3 + y^3 + z^3} \)
\( = \frac{3(x^2 + y^2 + z^2)}{x^3 + y^3 + z^3} \)
This type of problem helps understand how quickly a logarithmic function of several variables changes when all variables are altered simultaneously.
In simple words: We have a rule (U) that uses a logarithm and depends on x, y, and z. We find how this rule changes for x, then for y, then for z. Finally, we add up all these changes. The answer tells us the total rate of change if all three parts are changing.

🎯 Exam Tip: Remember the chain rule for logarithmic functions: the derivative of \( \log(f(x)) \) is \( \frac{f'(x)}{f(x)} \). Be careful to only differentiate with respect to the specified variable, treating others as constants.

Question 5. For each of the following functions find the \( g_{xy}, g_{xx}, g_{yy} \) and \( g_{yx} \).
(i) \( g(x, y) = x e^y + 3x^2y \)
(ii) \( g(x, y) = \log (5x + 3y) \)
(iii) \( g(x, y) = x^2 + 3xy – 7y + \cos(5x) \)
Answer:
(i) For \( g(x, y) = x e^y + 3x^2y \):
First, find \( g_x \):
\( g_x = \frac{\partial}{\partial x}(x e^y + 3x^2y) = e^y + 6xy \)
Next, find \( g_y \):
\( g_y = \frac{\partial}{\partial y}(x e^y + 3x^2y) = x e^y + 3x^2 \)

Now, find the second partial derivatives:
\( g_{xx} = \frac{\partial}{\partial x}(g_x) = \frac{\partial}{\partial x}(e^y + 6xy) = 6y \)
\( g_{yy} = \frac{\partial}{\partial y}(g_y) = \frac{\partial}{\partial y}(x e^y + 3x^2) = x e^y \)
\( g_{xy} = \frac{\partial}{\partial y}(g_x) = \frac{\partial}{\partial y}(e^y + 6xy) = e^y + 6x \)
\( g_{yx} = \frac{\partial}{\partial x}(g_y) = \frac{\partial}{\partial x}(x e^y + 3x^2) = e^y + 6x \)
As expected for a smooth function, \( g_{xy} = g_{yx} \).

(ii) For \( g(x, y) = \log (5x + 3y) \):
First, find \( g_x \):
\( g_x = \frac{\partial}{\partial x}(\log (5x + 3y)) = \frac{1}{5x+3y} \cdot 5 = \frac{5}{5x+3y} = 5(5x+3y)^{-1} \)
Next, find \( g_y \):
\( g_y = \frac{\partial}{\partial y}(\log (5x + 3y)) = \frac{1}{5x+3y} \cdot 3 = \frac{3}{5x+3y} = 3(5x+3y)^{-1} \)

Now, find the second partial derivatives:
\( g_{xx} = \frac{\partial}{\partial x}(g_x) = \frac{\partial}{\partial x}(5(5x+3y)^{-1}) = 5(-1)(5x+3y)^{-2}(5) = \frac{-25}{(5x+3y)^2} \)
\( g_{yy} = \frac{\partial}{\partial y}(g_y) = \frac{\partial}{\partial y}(3(5x+3y)^{-1}) = 3(-1)(5x+3y)^{-2}(3) = \frac{-9}{(5x+3y)^2} \)
\( g_{xy} = \frac{\partial}{\partial y}(g_x) = \frac{\partial}{\partial y}(5(5x+3y)^{-1}) = 5(-1)(5x+3y)^{-2}(3) = \frac{-15}{(5x+3y)^2} \)
\( g_{yx} = \frac{\partial}{\partial x}(g_y) = \frac{\partial}{\partial x}(3(5x+3y)^{-1}) = 3(-1)(5x+3y)^{-2}(5) = \frac{-15}{(5x+3y)^2} \)
Again, \( g_{xy} = g_{yx} \).

(iii) For \( g(x, y) = x^2 + 3xy – 7y + \cos(5x) \):
First, find \( g_x \):
\( g_x = \frac{\partial}{\partial x}(x^2 + 3xy – 7y + \cos(5x)) = 2x + 3y - \sin(5x) \cdot 5 = 2x + 3y - 5\sin(5x) \)
Next, find \( g_y \):
\( g_y = \frac{\partial}{\partial y}(x^2 + 3xy – 7y + \cos(5x)) = 0 + 3x - 7 + 0 = 3x - 7 \)

Now, find the second partial derivatives:
\( g_{xx} = \frac{\partial}{\partial x}(g_x) = \frac{\partial}{\partial x}(2x + 3y - 5\sin(5x)) = 2 - 5\cos(5x) \cdot 5 = 2 - 25\cos(5x) \)
\( g_{yy} = \frac{\partial}{\partial y}(g_y) = \frac{\partial}{\partial y}(3x - 7) = 0 \)
\( g_{xy} = \frac{\partial}{\partial y}(g_x) = \frac{\partial}{\partial y}(2x + 3y - 5\sin(5x)) = 3 \)
\( g_{yx} = \frac{\partial}{\partial x}(g_y) = \frac{\partial}{\partial x}(3x - 7) = 3 \)
Once more, \( g_{xy} = g_{yx} \). Understanding these derivatives helps in analyzing how the slope or rate of change of a surface behaves in different directions.
In simple words: We find how a function changes in different ways. First, we find how it changes along the x-direction (\( g_x \)) and along the y-direction (\( g_y \)). Then, we find the "change of the change." \( g_{xx} \) is how \( g_x \) changes in x; \( g_{yy} \) is how \( g_y \) changes in y. \( g_{xy} \) is how \( g_x \) changes in y, and \( g_{yx} \) is how \( g_y \) changes in x. For these functions, \( g_{xy} \) and \( g_{yx} \) are always the same.

🎯 Exam Tip: Pay close attention to which variable you are differentiating with respect to at each step. For mixed partials, always verify that \( g_{xy} = g_{yx} \), as this is a fundamental property of continuous functions.

Question 6. Let \( w(x, y, z) = \frac{1}{\sqrt{x^2+y^2+z^2}} \), \( (x, y, z) \neq (0, 0, 0) \), show that \( \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} = 0 \).
Answer:
Given \( w(x, y, z) = \frac{1}{\sqrt{x^2+y^2+z^2}} = (x^2+y^2+z^2)^{-1/2} \).

First, find \( \frac{\partial w}{\partial x} \):
Using the chain rule:
\( \frac{\partial w}{\partial x} = -\frac{1}{2}(x^2+y^2+z^2)^{-3/2} \cdot (2x) \)
\( = -x(x^2+y^2+z^2)^{-3/2} \)

Next, find \( \frac{\partial^2 w}{\partial x^2} \):
Using the product rule \( (uv)' = u'v + uv' \):
Let \( u = -x \implies u' = -1 \)
Let \( v = (x^2+y^2+z^2)^{-3/2} \implies v' = -\frac{3}{2}(x^2+y^2+z^2)^{-5/2} \cdot (2x) = -3x(x^2+y^2+z^2)^{-5/2} \)
\( \frac{\partial^2 w}{\partial x^2} = (-1)(x^2+y^2+z^2)^{-3/2} + (-x)(-3x(x^2+y^2+z^2)^{-5/2}) \)
\( = -(x^2+y^2+z^2)^{-3/2} + 3x^2(x^2+y^2+z^2)^{-5/2} \)
To combine these terms, factor out the common term \( (x^2+y^2+z^2)^{-5/2} \):
\( \frac{\partial^2 w}{\partial x^2} = (x^2+y^2+z^2)^{-5/2} [-(x^2+y^2+z^2) + 3x^2] \)
\( = (x^2+y^2+z^2)^{-5/2} [-x^2-y^2-z^2 + 3x^2] \)
\( = (x^2+y^2+z^2)^{-5/2} [2x^2-y^2-z^2] \)

By symmetry, we can find \( \frac{\partial^2 w}{\partial y^2} \) and \( \frac{\partial^2 w}{\partial z^2} \) by replacing \( x^2 \) with \( y^2 \) and \( z^2 \) respectively:
\( \frac{\partial^2 w}{\partial y^2} = (x^2+y^2+z^2)^{-5/2} [-x^2+2y^2-z^2] \)
\( \frac{\partial^2 w}{\partial z^2} = (x^2+y^2+z^2)^{-5/2} [-x^2-y^2+2z^2] \)

Now, sum these second partial derivatives:
\( \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} = (x^2+y^2+z^2)^{-5/2} [(2x^2-y^2-z^2) + (-x^2+2y^2-z^2) + (-x^2-y^2+2z^2)] \)
\( = (x^2+y^2+z^2)^{-5/2} [ (2x^2-x^2-x^2) + (-y^2+2y^2-y^2) + (-z^2-z^2+2z^2) ] \)
\( = (x^2+y^2+z^2)^{-5/2} [ 0 + 0 + 0 ] \)
\( = 0 \)
This result is significant as it shows that the function \( w \) satisfies Laplace's equation, meaning it is a harmonic function, which is important in physics and engineering.
In simple words: We have a special mathematical rule for 'w'. We find how 'w' changes twice for x, twice for y, and twice for z. When we add up all these "second changes," the total sum comes out to zero. This zero means the rule for 'w' is a special kind of "balanced" function.

🎯 Exam Tip: Remember to apply the chain rule and product rule correctly when finding the derivatives. Look for symmetry in the function to simplify calculations for other variables once one derivative is found.

Question 7. If \( V(x, y) = e^x (x \cos y – y \sin y) \), then Prove that \( \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0 \).
Answer:
Given \( V(x, y) = e^x (x \cos y – y \sin y) \).

First, find \( \frac{\partial V}{\partial x} \):
Using the product rule for \( (e^x)(x \cos y – y \sin y) \), treating \( (x \cos y – y \sin y) \) as a constant factor with respect to x.
\( \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(e^x) (x \cos y – y \sin y) + e^x \frac{\partial}{\partial x}(x \cos y – y \sin y) \)
\( = e^x (x \cos y – y \sin y) + e^x (\cos y – 0) \)
\( = e^x (x \cos y – y \sin y + \cos y) \)

Next, find \( \frac{\partial^2 V}{\partial x^2} \):
Apply the product rule again to \( e^x (x \cos y – y \sin y + \cos y) \).
\( \frac{\partial^2 V}{\partial x^2} = \frac{\partial}{\partial x}(e^x) (x \cos y – y \sin y + \cos y) + e^x \frac{\partial}{\partial x}(x \cos y – y \sin y + \cos y) \)
\( = e^x (x \cos y – y \sin y + \cos y) + e^x (\cos y – 0 – 0) \)
\( = e^x (x \cos y – y \sin y + 2 \cos y) \)

Now, find \( \frac{\partial V}{\partial y} \):
Treat \( e^x \) as a constant multiplier.
\( \frac{\partial V}{\partial y} = e^x \frac{\partial}{\partial y}(x \cos y – y \sin y) \)
Using the product rule for \( y \sin y \): \( \frac{\partial}{\partial y}(y \sin y) = (1)\sin y + y(\cos y) = \sin y + y \cos y \).
\( \frac{\partial V}{\partial y} = e^x (-x \sin y – (\sin y + y \cos y)) \)
\( = e^x (-x \sin y – \sin y – y \cos y) \)
\( = -e^x (x \sin y + \sin y + y \cos y) \)

Next, find \( \frac{\partial^2 V}{\partial y^2} \):
Treat \( -e^x \) as a constant multiplier.
\( \frac{\partial^2 V}{\partial y^2} = -e^x \frac{\partial}{\partial y}(x \sin y + \sin y + y \cos y) \)
Using the product rule for \( y \cos y \): \( \frac{\partial}{\partial y}(y \cos y) = (1)\cos y + y(-\sin y) = \cos y – y \sin y \).
\( \frac{\partial^2 V}{\partial y^2} = -e^x (x \cos y + \cos y + (\cos y – y \sin y)) \)
\( = -e^x (x \cos y + 2 \cos y – y \sin y) \)
\( = e^x (-x \cos y - 2 \cos y + y \sin y) \)

Finally, sum \( \frac{\partial^2 V}{\partial x^2} \) and \( \frac{\partial^2 V}{\partial y^2} \):
\( \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = e^x (x \cos y – y \sin y + 2 \cos y) + e^x (-x \cos y - 2 \cos y + y \sin y) \)
Factor out \( e^x \):
\( = e^x [ (x \cos y – y \sin y + 2 \cos y) + (-x \cos y - 2 \cos y + y \sin y) ] \)
\( = e^x [ (x \cos y - x \cos y) + (-y \sin y + y \sin y) + (2 \cos y - 2 \cos y) ] \)
\( = e^x [0 + 0 + 0] \)
\( = 0 \)
This proves that the function \( V \) also satisfies Laplace's equation, indicating its harmonic nature, which is fundamental in various areas of physics and engineering, especially for potentials and steady-state distributions.
In simple words: We have a rule for V that uses exponential and trigonometric parts. We find how this rule changes twice when only x changes, and then how it changes twice when only y changes. When we add these two "double changes" together, the answer is always zero. This shows a special balance in the rule.

🎯 Exam Tip: Be very careful with the product rule and chain rule, especially when differentiating terms like \( y \sin y \) with respect to y. Keep track of negative signs and exponential factors.

Question 8. If \( w(x, y) = xy + \sin (xy) \), then Prove that \( \frac{\partial^2 w}{\partial y \partial x} = \frac{\partial^2 w}{\partial x \partial y} \).
Answer:
Given \( w(x, y) = xy + \sin (xy) \).

First, find \( \frac{\partial w}{\partial x} \):
\( \frac{\partial w}{\partial x} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial x}(\sin (xy)) \)
\( = y(1) + \cos(xy) \cdot y \)
\( = y + y \cos(xy) \)

Next, find \( \frac{\partial^2 w}{\partial y \partial x} \): (This means differentiating \( \frac{\partial w}{\partial x} \) with respect to y)
\( \frac{\partial^2 w}{\partial y \partial x} = \frac{\partial}{\partial y}(y + y \cos(xy)) \)
Using the product rule for \( y \cos(xy) \): \( \frac{\partial}{\partial y}(y \cos(xy)) = (1)\cos(xy) + y(-\sin(xy) \cdot x) = \cos(xy) - xy \sin(xy) \).
\( \frac{\partial^2 w}{\partial y \partial x} = 1 + (\cos(xy) - xy \sin(xy)) \)
\( = 1 + \cos(xy) - xy \sin(xy) \) .....(1)

Now, find \( \frac{\partial w}{\partial y} \):
\( \frac{\partial w}{\partial y} = \frac{\partial}{\partial y}(xy) + \frac{\partial}{\partial y}(\sin (xy)) \)
\( = x(1) + \cos(xy) \cdot x \)
\( = x + x \cos(xy) \)

Next, find \( \frac{\partial^2 w}{\partial x \partial y} \): (This means differentiating \( \frac{\partial w}{\partial y} \) with respect to x)
\( \frac{\partial^2 w}{\partial x \partial y} = \frac{\partial}{\partial x}(x + x \cos(xy)) \)
Using the product rule for \( x \cos(xy) \): \( \frac{\partial}{\partial x}(x \cos(xy)) = (1)\cos(xy) + x(-\sin(xy) \cdot y) = \cos(xy) - xy \sin(xy) \).
\( \frac{\partial^2 w}{\partial x \partial y} = 1 + (\cos(xy) - xy \sin(xy)) \)
\( = 1 + \cos(xy) - xy \sin(xy) \) .....(2)

From (1) and (2), we can see that:
\( \frac{\partial^2 w}{\partial y \partial x} = \frac{\partial^2 w}{\partial x \partial y} \)
This verifies Clairaut's Theorem (also known as Schwarz's Theorem), which states that for a function with continuous second partial derivatives, the order of differentiation does not matter.
In simple words: We have a math rule for 'w' that depends on x and y. We find its change first for x, then that result's change for y. Then, we find its change first for y, and then that result's change for x. The answer shows that both ways lead to the exact same result, meaning the order of taking these "changes of changes" does not matter.

🎯 Exam Tip: When proving equality of mixed partial derivatives, carefully apply the product rule and chain rule, especially for trigonometric functions. Keep the steps clear and organized to avoid errors.

Question 9. If \( v(x, y, z) = x^3 + y^3 + z^3 + 3xyz \), Show that \( \frac{\partial^2 v}{\partial y \partial z} = \frac{\partial^2 v}{\partial z \partial y} \).
Answer:
Given \( v(x, y, z) = x^3 + y^3 + z^3 + 3xyz \).

First, find \( \frac{\partial v}{\partial z} \):
\( \frac{\partial v}{\partial z} = \frac{\partial}{\partial z}(x^3) + \frac{\partial}{\partial z}(y^3) + \frac{\partial}{\partial z}(z^3) + \frac{\partial}{\partial z}(3xyz) \)
\( = 0 + 0 + 3z^2 + 3xy(1) \)
\( = 3z^2 + 3xy \)

Next, find \( \frac{\partial^2 v}{\partial y \partial z} \): (This means differentiating \( \frac{\partial v}{\partial z} \) with respect to y)
\( \frac{\partial^2 v}{\partial y \partial z} = \frac{\partial}{\partial y}(3z^2 + 3xy) \)
\( = 0 + 3x(1) \)
\( = 3x \) .....(1)

Now, find \( \frac{\partial v}{\partial y} \):
\( \frac{\partial v}{\partial y} = \frac{\partial}{\partial y}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial y}(z^3) + \frac{\partial}{\partial y}(3xyz) \)
\( = 0 + 3y^2 + 0 + 3xz(1) \)
\( = 3y^2 + 3xz \)

Next, find \( \frac{\partial^2 v}{\partial z \partial y} \): (This means differentiating \( \frac{\partial v}{\partial y} \) with respect to z)
\( \frac{\partial^2 v}{\partial z \partial y} = \frac{\partial}{\partial z}(3y^2 + 3xz) \)
\( = 0 + 3x(1) \)
\( = 3x \) .....(2)

From (1) and (2), we can see that:
\( \frac{\partial^2 v}{\partial y \partial z} = \frac{\partial^2 v}{\partial z \partial y} \)
This again demonstrates that for a well-behaved function like this polynomial, the order of taking mixed partial derivatives does not affect the final result. This principle simplifies complex calculations in multi-variable calculus.
In simple words: We have a rule for 'v' that changes with x, y, and z. We first find how 'v' changes with z, and then how that result changes with y. Separately, we find how 'v' changes with y, and then how that result changes with z. Both ways, the final answer is the same, proving that the order of these two changes doesn't matter for this rule.

🎯 Exam Tip: Remember to treat other variables as constants during partial differentiation. For polynomial functions, derivatives are usually straightforward. The key is careful application of the rules and organization of steps.

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.
Answer:
(i) To find the Profit function \( P(x, y) \), we subtract the Cost function \( C(x, y) \) from the Revenue function \( R(x, y) \). The profit function helps businesses understand how their total profit changes based on the quantities of different products.
\( P(x, y) = R(x, y) - C(x, y) \)
\( P(x, y) = (80x + 90y + 0.04xy - 0.05x^2 - 0.05y^2) - (8x + 6y + 2000) \)
\( P(x, y) = 80x + 90y + 0.04xy - 0.05x^2 - 0.05y^2 - 8x - 6y - 2000 \)
Now, we combine the similar terms:
\( P(x, y) = (80x - 8x) + (90y - 6y) + 0.04xy - 0.05x^2 - 0.05y^2 - 2000 \)
\( P(x, y) = 72x + 84y + 0.04xy - 0.05x^2 - 0.05y^2 - 2000 \)

(ii) First, we find the partial derivative of the Profit function with respect to \( x \). This tells us how profit changes when the quantity of product A changes, while product B's quantity stays the same.
\( \frac{\partial P}{\partial x} = \frac{\partial}{\partial x} (72x + 84y + 0.04xy - 0.05x^2 - 0.05y^2 - 2000) \)
\( \frac{\partial P}{\partial x} = 72 + 0.04y - 0.1x \)
Next, we evaluate this at the given points \( x = 1200 \) and \( y = 1800 \):
\( \frac{\partial P}{\partial x} (1200, 1800) = 72 + 0.04(1800) - 0.1(1200) \)
\( \frac{\partial P}{\partial x} (1200, 1800) = 72 + 72 - 120 \)
\( \frac{\partial P}{\partial x} (1200, 1800) = 144 - 120 \)
\( \frac{\partial P}{\partial x} (1200, 1800) = 24 \)
This means if the company is currently producing 1200 units of type A and 1800 units of type B, producing one more unit of type A would increase the profit by Rs. 24.

Then, we find the partial derivative of the Profit function with respect to \( y \). This shows how profit changes when the quantity of product B changes, with product A's quantity remaining constant.
\( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (72x + 84y + 0.04xy - 0.05x^2 - 0.05y^2 - 2000) \)
\( \frac{\partial P}{\partial y} = 84 + 0.04x - 0.1y \)
Now, we evaluate this at \( x = 1200 \) and \( y = 1800 \):
\( \frac{\partial P}{\partial y} (1200, 1800) = 84 + 0.04(1200) - 0.1(1800) \)
\( \frac{\partial P}{\partial y} (1200, 1800) = 84 + 48 - 180 \)
\( \frac{\partial P}{\partial y} (1200, 1800) = 132 - 180 \)
\( \frac{\partial P}{\partial y} (1200, 1800) = -48 \)
This means if the company is currently producing 1200 units of type A and 1800 units of type B, producing one more unit of type B would decrease the profit by Rs. 48.
In simple words: First, we write down how much money is made (Revenue) and how much is spent (Cost). Subtracting these gives us the total Profit formula. Then, we look at how profit changes if we make one more of product A, keeping product B the same. We find it goes up by Rs. 24. Last, we look at how profit changes if we make one more of product B, keeping product A the same. We find it goes down by Rs. 48.

🎯 Exam Tip: When dealing with profit functions, make sure to simplify the expression carefully by combining like terms. For partial derivatives, remember to treat other variables as constants during differentiation and interpret the positive or negative sign correctly in your final answer.