Suppose that W(x, y) is a function of two variables x, y having partial derivatives \(\frac{\partial W}{\partial x}, \frac{\partial W}{\partial y}\) If both the variables x, y are differentiable functions of a single variable t , then W is a differentiable function of t and \(\frac{d W}{d t}=\frac{\partial W}{\partial x} \frac{d x}{d t}+\frac{\partial W}{\partial y} \frac{d y}{d t}\)

Given V (x, y, z) = xy + yz + zx, x, y, z ∈ R
dV = x.dx + y.dx + y.dz + z.dy + z.dx + x.dz
dV = (y+z)dx + (x+ z)dy + (y+x)dz

Givenz (x, y) = x²y+ 3xy⁴, x, y ∈ R.
z (x_o, y_o) = z (2, -1) = 2²(-1) + 3(2)(-1)4
= -4 + 6 = 2
\(\frac { \partial z }{ \partial x } \) = 2xy + 3y⁴
\(\left( \frac { \partial z }{ \partial x } \right) \)_{(2, -1)} = 2 (2) (-1) + 3 (1)4
= 4+3 = -1
\(\frac { \partial v }{ \partial y} \) = x² + 3x (4y³)
= x2 + 12xy3
\(\left( \frac { \partial z }{ \partial y } \right) \)_{(2, -1)} = 22+ 12 (2)(-1)³
= 4 -24 = -20
Linear approximate is given by
L(x, y) = z(x_o, y_o) + \(\left( \frac { \partial z }{ \partial x} \right) \)(x_o, y_o) (x - x₀) 
L(x, y) = z(x_o, y_o) + \(\left( \frac { \partial z }{ \partial z} \right) \)(x_o, y_o) (x - x₀)
∴ L(x, y) = 2-1 (x-2) - 20(y + 1)
= 2 - x + 2 - 20y - 20
= -x - 20y - 16
= - (x + 20y + 16)

Given w (x, y) = x³ - 3xy+ 2.0, x, y ∊ R
w_x = \(\frac { \partial w }{ \partial x } \) = 3x² - 3y + 0 = 3x² - 3y 
w_x = at (1, -1)= 3 (1p - 3- (-1) = 3 + 3 = 6
wy = \(\frac { \partial w }{ \partial y } \) = 0 -3x+4y = -3x+4y
w_y (1, -1) = -3 (1)+4(-1)= -3 - 4 = -7.
w (x_o, y_o) = w (1, -1)
= 1³- 3(1) (-1) + 2 (-1)²
= 1+3+2 = 6
Linear approximation is given by
L(x,y) = w(x_o,y_o) + \((\frac { \partial w }{ \partial x })_{(x_o,y_o) }\)(x-x₀) + \((\frac { \partial w }{ \partial xy})_{(x_o,y_o) }\) (y-y_o)
= 6 + 6 (x - 1) + -7 (y + 1)
= 6 + 6x - 6 - 7y - 7
= 6x - 7y - 7

By (14), Linear approximation is given by
L (x, y, z) =  U(x₀, y₀, z₀) + \({ \frac { { \partial }U }{ { \partial x } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (x-x₀)+\({ \frac { { \partial }U }{ { \partial y } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (y-y₀)+\({ \frac { { \partial }U }{ { \partial z } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (z-z₀)
Now U_x = 2x -y, Uy = -xand U_z = 3cos z.
Here (x₀, y₀, z₀) = (2,−1,0 )
hence U_x (2, −1,0) = 5, U_y (2, −1,0) = −2 and U_z (2,-1,0) = 3.
Thus L(x, y, z) = 6 + 5(x − 2) − 2( y +1) + 3(z − 0) = 5x − 2y + 3z − 6 is the required linear approximation for U at (2,−1,0).

Given g(x, y) = \(\frac { { e }^{ y }sinx }{ x } \) for x ≠ 0 and g(0, 0) = 1
g(0, 0) = 1
The function g is defined for all (x, y) ∈ R²
To check if g has a limit L at (0,0) and if L= g(0,0) = 1
Consider \(\left| g(x,y)-g(0,y) \right| =\left| \frac { { e }^{ y }sinx }{ x } -0 \right| \)
= \(\left| \frac { { e }^{ y }sinx }{ x } \right| =\left| \frac { \left| { e }^{ y } \right| \left| sinx \right| }{ x } \right| =\left| { e }^{ x } \right| \left| \frac { sinx }{ x } \right| =1\)
\(\left[ \because (x,y)\longrightarrow (0,0)\Rightarrow \left| { e }^{ y } \right| =1and\left| \frac { sinx }{ x } \right| =1 \right] \)
\(\because \begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}\frac { { e }^{ y }sinx }{ x } =1=g(0,0)\) Which proves that is continuous at (0, 0)
∴ g(x, y) is continuous at (0, 0)

Let (a, b)  ∈ R² an arbitrary point we shall investigate continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a, b).
(i) f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) is defined
(ii) \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}=\frac { \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ x }^{ 2 }-{ y }^{ 2 } }{ \begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}{ y }^{ 2 }+1 } \) = L exists
= \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \) = L exists
(iii) Also, f(a, b) = \(\frac { { a }^{ 2 }-{ b }^{ 2 } }{ b^{ 2 }+1 } \)
∴ \(\begin{matrix} lim \\ (x,y)\rightarrow (a,b) \end{matrix}\)= L = f(a, b)
Hence f satisfies all the three conditions since (a, b) is an arbitrary point on R², we conclude that f is continuous at every point of R².

Note that f is defined for every (x, y)∈R². First let us check the continuity at (a, b) ≠ (0, 0).
Let us say, just for instance, (a, b) = (2, 5). Then f(2, 5) = \(\frac{10}{29}\). Then, as in the above example, we callculate \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) xy = 2(5) and \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \)  x²+y² = 2²+5² = 29 ≠ 0.
Hence,  \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) = \(\frac { 10 }{ 29 } \).
Since f(2,5) = \(\frac { 10 }{ 29 } \) \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) it follows that f is continuous at (2, 5)
Exactly by similar arguments we can show that f is continuous at every point (a, b) ≠ (0, 0). Now let us check the continuity at (0, 0). Note that f (0, 0) = 0 by definition. Next we want to find if \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) exists or not.
First let us check the limit along the straight lines y = mx , passing through (0,0) .
 \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \)  = \(\underset { x\longrightarrow 0 }{ lim } \) \(\frac { m{ x }^{ 2 } }{ \left( 1+{ m }^{ 2 } \right) { x }^{ 2 } } =\frac { m }{ 1+{ m }^{ 2 } } \neq \) f (0, 0), if m ≠0.
So for different values of m, we get different values \(\frac { m }{ 1+{ m }^{ 2 } } \) and hence we conclude that   \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) does not exist. Hence f cannot be continuous at (0, 0).

Let (a,b)∈R² be an arbitrary point. We shall investigate continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a,b).
To check first condition, note that f (a, b) = \(\frac { 3x-5y+8 }{ { x }^{ 2 }+{ y }^{ 2 }+1 } \) is defined.
Next we want to find if \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) exists or not.
So we calculate (a,b) \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) (3x -5y +8) = 3a-5b+8 and \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) (x² +y²+1) = a²+b²+1 ≠ 0
Thus, by the properties of limits, we see that
\(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) f(x, y) = \(\frac { \underset { (x,y)\longrightarrow (a,b) }{ lim } \left( 3x-5y+8 \right) }{ \underset { (x,y)\longrightarrow (a,b) }{ lim } \left( { x }^{ 2 }+{ y }^{ 2 }+1 \right) } \) = \(\frac { 3a-5b+8 }{ { a }^{ 2 }+{ b }^{ 2 }+1 } \) = f(a, b) = L exists.
Now we note that \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) f(x, y) = L = f(a, b). Hence f satisfies all the three conditions for continuity of f at (a, b). Since (a, b) is an arbitrary point in R², we conclude that f is continuous at every point of R².

log₁₀e = 0.4343 to find log₁₀g 1003
f(1000) = log₁₀1000 = log₁₀10³ = 3log₁₀10³ = 3 log₁₀10
= 3(1) = 3
f'(x) = \(\frac1x\). log10^e
f'(1000) = \(\frac{1}{1000}\)(0.4343)
∴ L(x) = f(x₀) f'(x₀) (x - x₀)
= 3 + \(\frac{1}{1000}\) (0.4343) (3)
= 3 + \(\frac{1.3029}{1000}\)
= 3 + 0.0013029
log₁₀1003 = 3.0013029

Given f(x) = x³ - 2x² ; x = 2, ∆ x = dx = 0.5
(1) df = f'(x).∆x = (3.x² - 4x) ∆x
= [3(2)2 - 4(2)] (0.5)
= 4(0.5) = 2.0
∆f = f(x + ∆x) - f(x)
= f(2.5) - f(2)
= [(2.5)³ - 2 (2.5)²] - [2³ - 2(2²)]
= 15.625 -12.5 -0 = 3.125
(2) df = f'(x) ∆x (2x + 2) (∆x)
x = -0.5, ∆x = dx = 0.1
= (2(-0.05) + 2) = 0.1
∆f = f(x + ∆x) - f(x)
= f(-0.5 + 0.1) - f(-0.5)
= f(0.4) - f(-0.5)
= [(-0.4)² + 2(-0.4) + 3] - [(-0.5)² + 2(-0.5) + 3]
= (16 - 0.8 + 3) - (0.25 - 1 + 3)
= 2.36 - 2.25 = 0.11 

We know that volume of a sphere is given by V = \(\frac43\) π r³. where r > 0 is the radius. So the differential dV = 4 ㅠr² dr and hence
Δ ≈ dv = 4π(10)² (9.9-10) cm³
= 4π10² (-0.1) cm³
= −40π cm³ 
Note that we have used dr = (9.9 −10) cm, because radius decreases from 10 to 9.9. Again the negative sign in the answer indicates that the volume of the sphere decreases about 40π cm³.

Let f, g : (a, b)→R be differentiable functions and h(x) = f (x)g(x).
Then h being product differentiable functions, is differentiable on (a,b) 
So by definition dh = h'(x)dx.
Now by using product rule we have h'(x) = f (x)g'(x) + f'(x)g(x).
Thus dh = h'(x)dx = ( f (x)g'(x) + f''(x)g(x))dx = f (x)g'(x)dx + f '(x)g(x) dx
= f (x)dg + g(x)df = fdg + gdf

\({ h }({ x }_{ o })=\frac { x }{ 1+1 } =\frac { 1 }{ 2 } \)
\({ h }^{ ' }(x)=\frac { (x+1)(1)-x(1) }{ { (x+1) }^{ 2 } } \)
\(\frac { x+1-x }{ { (x+1) }^{ 2 } } =\frac { 1 }{ ({ x+1) }^{ 2 } } \)
\({ h }^{ ' }({ x }_{ o })=\frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ 4 } \)
∴ L(x) = h(x_o) + h'(x₀)(x - x_o)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } (x-1)=\frac { 2+x-1 }{ 4 } =\frac { x+1 }{ 4 } \)
∴ L(x) = \(\frac { x+1 }{ 4 } \)

Given \(g(x)=\sqrt { { x }^{ 2 }+9 } ,{ x }_{ 0 }=-4\)
\(g(x)=\sqrt { { (-4) }^{ 2 }+9 } =\sqrt { 16+9 } =5\)
\({ g }^{ ' }(x)=\frac { 1 }{ 2 } ({ { x }^{ 2 }+9 })^{ -\frac { 1 }{ 2 } }(2x)=\frac { x }{ \sqrt { { x }^{ 2 }+9 } } \)
\(\therefore { g }^{ ' }({ x }_{ 0 })=\frac { -4 }{ \sqrt { { (-4) }^{ 2 }+9 } } =\frac { -4 }{ 5 } \)
∴ L(x) = g(x_o) + g'(x₀)(x - x_o)
= \(5-\frac { 4 }{ 5 } (x+4)=\frac { 25-4x-16 }{ 5 } \)
L(x) = \(\frac { 9-4x }{ 5 } \)

f(x) = x³ - 5x + 12, x₀ = 2
f(x_o) = 2³ - 5(2) + 12
= 8 - 10 + 12 = 10
f'(x) = 3x² - 5
⇒ f'(x_o) = 3 (2²) - 5 = 7
∴ L(x) = f(x_o) +f'(x_o) (x - x_o)
= 10 + 7(x - 2)
= 10 + 7x - 14
L(x) = 7x- 4 

Given \(f(x)=\sqrt [ 3 ]{ x } \)
Let x₀ = 27 and \(\triangle x=0.2\)
We know L(x) = \(f({ x }_{ 0 })+{ f }^{ ' }({ x }_{ o })\) (x - x₀) ∀ x ∈ (a, b)
∴ \(\sqrt [ 3 ]{ 27.2 } =f(27)+{ f }^{ ' }(27)(0.2)\) .. (1)
Now f(27) = \(\sqrt [ 3 ]{ 27 } =3\)
\({ f }^{ ' }(27)=\frac { 1 }{ 3 } x^{ \frac { 1 }{ 3 } -1 }=\frac { 1 }{ 3 } { x }^{ \frac { 2 }{ 3 } }=\frac { 1 }{ { 3x }^{ \frac { 2 }{ 3 } } } \)
\(\therefore \) becomes,
\(\sqrt [ 3 ]{ 27.2 } =3+\frac { 1 }{ 27 } (0.2)\)
= 3 + .0074 = 3.0074
\(\therefore \) \(\sqrt [ 3 ]{ 27.2 } =3.0074\)

Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

We need to find an approximate value of \(\sqrt { 9.2 } \) using linear approximation. Now by (3), we have f(x₀+Δx) ≈ f(x₀)+f'(x₀)Δx. To do this, we have to identify an appropriate function f, a point x₀ and Δx. Our choice should be such that the right side of the above approximate equality, should be computable without the help of a calculator. So, we choose  
f(x) = \(\sqrt { x,{ x }_{ 0 } } \) = 9  and Δx = 0.2. Then f'(x₀) = \(\frac { 1 }{ 2\sqrt { 9 } } \) and hence.
 \(\sqrt { 9.2 } \) ≈ f(9) + f'(9)(0.2) = 3+\(\frac { 0.2 }{ 6 } \) = 3.03333
Now if we use a calculator, just to compare, we find \(\sqrt { 9.2 } \) = 3.03315. We see that our approximation is accurate to three decimal places and the error is 3.03315 - 3.03333 = 0.00018. [Also note that one could choose f (x) = \(\sqrt { 1+x,{ x }_{ 0 } } =8\) and Δx = 0.2. So the choice of f and x₀ - are not necessarily unique].
 So in the above example, the absolute error is 3.03315-3.03333 = -0.00018. Note that the absolute error says how much the error; but it does not say how good the approximation is. For instance, let us consider two simple cases
Case 1 : Suppose that the actual value of something is 5 and its approximated value is 4, then the absolute error is 5 − 4 = 1.
Case 2 : Suppose that the actual value of something is 100 and its approximated value is 95. In this case, the absolute error is 100 − 95 = 5. So the absolute error in the first case is smaller when compared to the second case.
Among these two approximations, which is a better approximation; and why? The absolute error does not give a clear picture about whether an approximation is a good one or not. On the other hand, if we calculate relative error or percentage of error (defined below), it will be easy to see how good an approximation is. If the actual value is zero, then we do know how close our approximate answer is to the actual value. So if the actual value is not zero.

We know from (4), that L(x) = f(x₀) +f'(x₀)(x-x₀) We have x₀ = 3, \(\Delta \)x = 0.2 and hence f(3) = \(\sqrt { 1+3 } \) = 2. Also,
f′(x) = \(\frac { 1 }{ 2\sqrt { 1+x } } \) and hence f'(3) = \(\frac { 1 }{ 2\sqrt { 1+3 } } \) = \(\frac { 1 }{ 4 } \)
Thus, L(x) = 2 +\(\frac { 1 }{ 4 } \)(x-3) = \(\frac { x }{ 4 } \)+\(\frac { 5 }{ 4 } \) gives the required linear approximation.
Now, f(3.2) = \(\sqrt { 4.2 } \) ≈ L(3.2) = \(\frac { 3.2 }{ 4 } \)+\(\frac { 5 }{ 4 } \) = 2.050
Actually, if we use a calculator to calculate we get \(\sqrt { 4.2 } \) = 2.04939

\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}cos=\left( \frac { { x }^{ 3 }+{ y }^{ 3 } }{ x+y+2 } \right) \)
\(=cos\left( \frac { 0+0 }{ 0+0+2 } \right) =cos0=1\)

Given g(x, y) = \(\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}g(x,y)=\begin{matrix} lim \\ (x,y)\rightarrow (1,2) \end{matrix}\frac { { 3x }^{ 2 }-xy }{ { x }^{ 2 }+{ y }^{ 2 }+3 } \)
\(=\frac { { 3(1) }^{ 2 }-1(2) }{ { 1 }^{ 2 }+{ 2 }^{ 2 }+3 } =\frac { 3-2 }{ 8 } =\frac { 1 }{ 8 } \)

Given r = 2 mm
dr = (2.1 - 2) = 0.1 mm
Area = πr²
Approximate area dA = 2πr dr
= 2π (2) (0.1)
= 4 π (0.1) = 0.4 π mm²

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

x = 2 and dx = 0.1
Taking differentials,
df = (2x + 3) dx
Whenx = 2, dx = 0.1
df = (2(2) + 3)(0.1) = 7(0.1) = 0.7

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

Given w(x, y, z) = \(\left( \frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
Let (x, y, z) = \(\frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
⇒ w = log f  ...(1)
⇒ e^w = f
f(λx, λy, λz) = \(\frac { { 5\lambda }^{ 3 }{ x }^{ 3 }{ \lambda }^{ 4 }{ y }^{ 4 }+7{ \lambda }^{ 2 }{ y }^{ 2 }\lambda x{ \lambda }^{ 4 }{ z }^{ 4 }-75{ \lambda }^{ 3 }{ y }^{ 3 }{ \lambda }^{ 4 }{ z }^{ 4 }{ }^{ } }{ { \lambda }^{ 2 }{ x }^{ 2 }+{ \lambda }^{ 2 }{ y }^{ 2 } } \)
= \(\frac { { \lambda }^{ 7 }(5{ x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75 }y^{ 3 }{ z }^{ 4 } }{ { \lambda }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) } ={ \lambda }^{ 5 }f(x,y,z)\)
∴ f(x, y, z) is a homogeneous function of degree 5.
∴ By Euler's theorem,
\(x.\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } +z\frac { \partial f }{ \partial z } =5.f\)
⇒ \(x.\frac { \partial }{ \partial x } ({ e }^{ w })+y.\frac { \partial }{ \partial y } ({ e }^{ w })+z.\frac { \partial }{ \partial z } ({ e }^{ w })=5.{ e }^{ w }\) [using (1)]
⇒ \(x.{ e }^{ w }\frac { \partial w }{ \partial x } +y.{ e }^{ w }\frac { \partial w }{ \partial y } +z.{ e }^{ w }\frac { \partial w }{ \partial z } ({ e }^{ w })=5{ e }^{ w }\)
⇒ \(x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } \)  [Divided by e^w]

Given v (x, y) = log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \)
Since log \(\left( \frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \right) \) is not homogeneous,
left f (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ x+y } \)
⇒ v = log f
⇒ ev = f
Now, f(tx, ty) = \(\frac { { t }^{ 2 }{ x }^{ 2 }+{ t }^{ 2 }{ y }^{ 2 } }{ tx+ty } =\frac { { t }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ t(x+y) } \)
= t.f(x, y)
∴ f is a homogeneous function of degree 1.
∴ By Euler's theorem,
⇒ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 1.f
From (1), \(x.\frac { \partial }{ \partial x } \left( { e }^{ v } \right) +y.\frac { \partial }{ \partial y } \left( { e }^{ v } \right) ={ e }^{ v }\) [using (1)]
⇒ \(x.{ e }^{ v }.\frac { \partial }{ \partial x } +y.{ e }^{ v }.\frac { \partial }{ \partial y } ={ e }^{ v }\)
⇒  \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = 1 [Divided by e^v]

Given g (x, y) = x log \(\left( \frac { y }{ x } \right) \)
g(λx, λy) = λx log \(\left( \frac { \lambda y }{ \lambda x } \right) \)
= λx log \(\left( \frac { \lambda y }{ \lambda x } \right) \)
= λ x log \(\left( \frac { y }{ x } \right) \)
= λ¹ g (x, y)
∴ g (x, y) is a homogeneous function of degree 1.
To verify \(x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } =1g\) [Euler's theorem]
∴ \(\frac { \partial g }{ \partial x } \) = \(x.\frac { 1 }{ \frac { y }{ x } } \left( -\frac { y }{ { x }^{ 2 } } \right) +log\left( \frac { y }{ x } \right) \)(1)
= \(-\frac { { x }^{ 2 } }{ y } \left( \frac { y }{ { x }^{ 2 } } \right) log\left( \frac { y }{ x } \right) \)
= - 1 + log \(\left( \frac { y }{ x } \right) \)
\(\frac { \partial g }{ \partial y } =x.\frac { 1 }{ \left( \frac { y }{ x } \right) } \left( \frac { 1 }{ x } \right) =\frac { 1 }{ \frac { y }{ x } } =\frac { x }{ y } \)
Consider \(x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } \)
= \(x\left( -1+log\frac { y }{ x } \right) +y\left( \frac { x }{ y } \right) \)

= x log \(\left( \frac { y }{ x } \right) \) = g
\(\therefore x\frac { \partial g }{ \partial x } +y\frac { \partial g }{ \partial y } \) = 1(g)
Hence, Euler's theorem is verified.

Given (x,y) = x³ - 2x²y + 3xy² + y³ ...(1)
f(tx, ty) = (tx)³ - 2(tx)² (ty) + 3 (tx) (ty)² + (ty)³
= t³ x³ - 2t² x²ty + 3txt²y²+ t³y³
= t³ (x³ - 2x²y + 3xy² +y³)
f(tx, ty) = t³.f(x, y)
∴ f is a homogeneous function and its degree is 3.
Differentiate (1) partially with respect to 'x' and 'y' we get
\(\frac { \partial f }{ \partial x } { =3 }^{ 2 }-4xy+3{ y }^{ 2 }\)
\(\Rightarrow x\frac { \partial f }{ \partial x } ={ 3x }^{ 2 }-4xy+{ 3xy }^{ 2 }\)  ...(2)
\(\frac { \partial f }{ \partial y } =-{ 2x }^{ 2 }+6xy+3{ xy }^{ 2 }\)
\(\Rightarrow y\frac { \partial f }{ \partial y } =-2{ x }^{ 2 }+{ 6xy }^{ 2 }+{ 3y }^{ 3 }\) ...(3)
Adding (2) and (3) we get,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3x² - 4x²y + 3x²y- 2x²y + 6xy² + 3y³
= 3x² - 6x²y + 9xy² + 3y³
= 3 (x³ - 2x²y - 3xy² +y³)
= 3f [using (1)]
∴ \(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } \) = 3f = nf where 3 is the degree of (x, y)
Hence Euler's theorem is verified.

Given z(x, y) = x³ - 3x²y³ ; x = se^t; y = se^-t
\(\frac { \partial z }{ \partial s } \) = 3x² - 6xy³; \(\frac { \partial z }{ \partial y } \) = -9x²y²
∴ \(\frac { \partial z }{ \partial s } \) 3s²e^2t - 6 se^t s³ e^-3t
\(\frac { \partial z }{ \partial y } \) = 9s²e^2t.s²e^-2t
\(\frac { \partial z }{ \partial y } \) = -9s⁴
\(\frac { \partial z }{ \partial s} \) = e^t; \(\frac { \partial y }{ \partial s} \) = e^-t
\(\frac { \partial x }{ \partial t} \) = se^t; \(\frac { \partial y }{ \partial s} \) = -se^-t 
By chain rule;
\(\therefore \frac { \partial z }{ \partial s } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial u }{ \partial y } .\frac { dy }{ dx } \)
= (3s²e^2t - 6 s⁴ e^-2t) (e^t) + (-9s⁴) (e^-t)
= 3s²e^3t - 6 s⁴ e^-t -9s⁴ e^-t
\(\frac { \partial z }{ \partial s } \) = 3s²e^3t - 15 s⁴ e^-t
By chain rule;
\(\frac { \partial z }{ \partial t } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
= (3s²e^2t - 6 s⁴e-^2t) (se^t) + (-9 s⁴) (-se^-t)
= 3s³e^3t - 6 s⁵e^-t + 9 s⁵ e^-t
= 3s³ e^3t + 3 s⁵e^-t
= 3s³(e^3t + s²e^-t)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

\(\frac { \partial z }{ \partial x } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } (y)+{ tan }^{ -1 }(xy)\)
\(\frac { \partial z }{ \partial y } =x.\frac { 1 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)(x)
\(\frac { \partial z }{ \partial z } =\frac { xy }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \) + tan^-1 (xy)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial y } =\frac { x^2 }{ 1+{ x }^{ 2 }{ y }^{ 2 } } \)
\(\frac { \partial z }{ \partial x } =\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } +{ tan }^{ -1 }({ t }^{ 2 }{ se }^{ 2 })\)
\(\frac { \partial z }{ \partial y } =\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\frac { dx }{ dt } =2t;\frac { dy }{ dt } ={ s.e }^{ t }\)
Also, \(\frac { dx }{ ds } =0;\frac { dy }{ ds } ={ e }^{ t }\)
By chain rule
\(\frac { dw }{ ds } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial z }{ \partial y } .\frac { dy }{ ds } \)
= \(\frac { { t }^{ 2 }{ se }^{ t } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } (0)+\frac { { t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } { (e }^{ t })=\frac { { e }^{ t }{ t }^{ 4 } }{ 1+{ t }^{ 4 }{ s }^{ 2 }{ e }^{ 2t } } \)
\(\therefore { \left( \frac { \partial z }{ \partial s } \right) }_{ s=t=1 }=\frac { e(1) }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
By chain rule
= \(\frac { dz }{ dt } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
\(\therefore { \left( \frac { \partial z }{ \partial t } \right) }_{ (s=t=1) }=\frac { e }{ 1+{ e }^{ 2 } } =\frac { e }{ 1+{ e }^{ 2 } } \)
= \(\frac { 2e+e }{ 1+{ e }^{ 2 } } =\frac { 3e }{ 1+{ e }^{ 2 } } \)+ 2 tan^-1(e)

Given u (x, y, z) = xyz; x = e^-t y = e^-t cas t; z = sin t
\(\frac { \partial u }{ \partial x } \) = yz; \(\frac { \partial u }{ \partial y } \) = xz; \(\frac { \partial u }{ \partial z } \) = xy
⇒ \(\frac { \partial u }{ \partial x } \) = e^t cas t sin t
\(\frac { \partial u }{ \partial y } \) = e^t sin t
\(\frac { \partial u }{ \partial z } \) = e^-2t cas t and
\(\frac{dx}{dt}=-e^{-t}\)
⇒ \(\frac{dy}{dt}\) = e^-t (- sin t) - cas t e^-t
⇒ \(\frac{dz}{dt}\) = cos t
∴ By chain rule;
\(\frac { dw }{ dt } =\frac { \partial w }{ \partial x } .\frac { dx }{ dt } +\frac { \partial w }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } +\frac { \partial u }{ \partial z } .\frac { dz }{ dt } \)
= e^-1 cos t sin t (-e^-t) + e^-t sin t (-e^-t) + e^-t sin t (-e^-t sin t - e^-t cos t)+ (-e-^2t) cos t(cos t)
= -e^-2t [(sin t cos t + sin² t + sin t cos t - cos² t]
= -e^-2t [2 sin t cos t - (cos² t - sin² t)]
\(\frac { du }{ dt } \) = - e^-2t, [sin 2t - cos2t]
[∵ cas 2 t = cos² t - sin² t and sin 2t = 2 sin t cos t]

Here again we shall use the tree diagram to calculate \(\frac { \partial g }{ \partial r } ,\frac { \partial g }{ \partial s } \) 
Hence we find \(\frac { \partial g }{ \partial x } \) = 2x, \(\frac { \partial g }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial y } \) = 2, \(\frac { \partial x }{ \partial s } \) = -1, \(\frac { \partial y }{ \partial r } \) =2r, and \(\frac { \partial y }{ \partial s } \) = 2.
Now, \(\frac { \partial g }{ \partial r } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial r } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial r } \) = 2x(2) + 2(2r) 12r - 4s.
also,  \(\frac { \partial g }{ \partial s } \) = \(\frac { \partial g }{ \partial x } \frac { \partial x }{ \partial s } +\frac { \partial g }{ \partial y } \frac { \partial y }{ \partial s } \) = 2x(-1) +(2)2 = 2s - 4r + 4.

We shall follow the tree diagram to calculate
So first we need to find \(\frac { \partial g }{ \partial x } ,\frac { \partial g }{ \partial y } ,\frac { dx }{ dt } \) and \(\frac { dx }{ dt } \).
Now, \(\frac { \partial g }{ \partial x } \) = 2x-y + cos(x + y), \(\frac { \partial g }{ \partial x } \) = -x+cos(x + y), \(\frac { dx }{ dt } \) = 3e^3t and \(\frac { dx }{ dt } \) = 2t.
Thus, \(\frac { dg }{ dt } =\frac { \partial g }{ \partial x } \frac { dx }{ dt } +\frac { \partial g }{ \partial y } \frac { dy }{ dt } \)
= (2x − y + cos(x + y)) 3e^3t + (−x + cos(x + y))( 2t)
= (2e^3t - t² + cos(e^3t - t²))3e^3t +( -e^3t + cos(e^3t - t²))(2t ) 
= 6e^6t - 3t² e^3t +3e^3t cos(e^3t - t²) -2te^3t +2t cos(e^3t - t²) 
Also, some times our W(x, y) will be such that x = x(s, t) , and y = y(s, t) where s, t ∈ R. Then W can be considered as a function that depends on s and t. If x, y both have partial derivatives with respect to s, t and W has partial derivatives with respect to x and y, then we can calculate the partial derivatives of W with respect to s and t using the following theorem.

Let F(x, y) = x² – 2y² + 2xy and x(t) = cost, y(t) = sint
Then F(x, y) = cos² t - 2sin² t + 2cos t sin t  and thus F has becomes a function of one variable t. So by using chain rule, we see that
\(\frac { dF }{ dt } \) = 2 cos t(-sin t) -4 sin t cos t 2 (-sin² t + cos² t).
= -6 cos t sin t +2 ( -sin² t + cos² t)
On the other hand if we calculate
\(\frac { \partial F }{ \partial x } \frac { \partial x }{ \partial t } +\frac { \partial F }{ \partial y } \frac { \partial y }{ \partial t } \) = (2x+2y)\(\frac { d x }{ dt } \)+(2x - 4y) \(\frac { dy }{ dt } \)
= 2(cos t + sin t)(-sin t) + 2(cos t - 2sin t)(cos t)
= -6 cos t sin t +2 ( -sin² t + cos² t)
= \(\frac { dF }{ dt } \)

Given g (x, y) = x² + 3xy - 7y + cos(5x)
g_x = 2x + 3y - 0 - 5 sin 5x
= 2x + 3y - 5 sin 5x
g_y = 0 + 3x (1) - 7 + 0
= 3x - 7
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\) = 3
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })\)
= 2(1) + 0 - 5(5) cos (5x)
= 2 - 25 cos (5x)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\) 0
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })\)
= 0 + 3(1) - 0 = 3

g(x, y) = log (5x + 3y)
\({ g }_{ x }=\frac { 1 }{ 5x+3y } (5)=\frac { 5 }{ 5x+3y } \)
\({ g }_{ y }=\frac { 1 }{ 5x+3y } (3)=\frac { 3 }{ 5x+3y } \)
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })\)
= 3(-1)(5x + 3y)^-2 (5)
\(=\frac { -15 }{ { (5x+3y) }^{ 2 } } \)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ x })=\frac { -5 }{ ({ 5x+3y) }^{ 2 } } (5)\)
\(=\frac { -25 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })=\frac { -3 }{ ({ 5x+3y) }^{ 2 } } (3)\)
\(=\frac { -9 }{ ({ 5x+3y) }^{ 2 } } \)
\({ g }_{ yx }=\frac { \partial }{ \partial y } ({ g }_{ x })=\frac { -5 }{ 5x+3y } (3)\)
\(=\frac { -15 }{ ({ 5x+3y) }^{ 2 } } \)

g(x, y) = xe^y + 3x²y
g_x = e^y + 6xy
gy = xey + 3x²
\({ g }_{ xy }=\frac { \partial }{ \partial x } ({ g }_{ y })={ e }^{ y }+6x\)
\({ g }_{ xx }=\frac { \partial }{ \partial x } ({ g }_{ xy })=0+6y=6y\)
\({ g }_{ yy }=\frac { \partial }{ \partial y } ({ g }_{ y })={ xe }^{ y }\)
\({ g }_{ xy }=\frac { \partial }{ \partial y } ({ g }_{ x })\) = e^y + 6

Given U(x, y, z) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy } +3{ z }^{ 2 }y\)
\(\frac { \partial U }{ \partial x } =\frac { xy(2x)-({ x }^{ 2 }+{ y }^{ 2 })(y) }{ { x }^{ 2 }{ y }^{ 2 } } +0\)
\(=\frac { 2{ x }^{ 2 }y-{ x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { y({ x }^{ 2 }-{ y }^{ 2 }) }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }y } \)
\(\frac { \partial U }{ \partial y } =\frac { xy(2y)-({ x }^{ 2 }+{ y }^{ 2 })(x) }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
= \(\frac { { 2xy }^{ 2 }-{ x }^{ 3 }-{ xy }^{ 2 } }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { xy }^{ 2 }-{ x }^{ 3} }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { y }^{ 2 }-{ x }^{ 2 } }{ x{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(\frac { \partial U }{ \partial z } =0+3y(2z)=6yz\)

Given f(x, y) = cos (x² - 3xy)
f_x = - sin (x² - 3xy) [2x - 3y]
= (3y - 2x) sin (x² - 3xy)
f_y = - sin (x² - 3xy) (-3x)
= 3x sin (x² - 3xy)
∴ f_xy = \(\frac { \partial }{ \partial x } ({ f }_{ y })\)
= 3 [x cos (x² - 3xy) (2x - 3y) + sin (x² - 3xy) (1)]
= 3[(2x²- 3xy)cos(x² - 3xy) + sin (x² - 3xy)] ... (1)
∴ f_yx  = \(\frac { \partial }{ \partial y} ({ f }_{ x })\)
= (3y- 2x) cos(x² - 3xy) (- 3x) + sin (x² - 3xy) (3)
= 3[(2x²- 3xy)cos(x² - 3xy) + sin (x² - 3xy)] ... (2)
From (1) and (2),f_xy =1.

\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)  ...(2)
∴ From (1) and (2), f_xy = f_yz

Given f(x, y) = \(\frac { 3x }{ y+sinx \ } \)
\({ f }_{ x }=\frac { \partial f }{ \partial x } \)
= \(\frac { (y+sin \ x)(3)-3x(cos \ x) }{ { (y+sin \ x) }^{ 2 } } \)
\(=\frac { 3y+3sin \ x-3x \ cos \ x }{ (y+sin{ x })^{ 2 } } \)
\({ f }_{ yx }=\frac { \partial ^{ 2 }f }{ \partial y\partial x } \)
\(=\frac { (y+{ sinx) }^{ 2 }[3]-(3y+3sinx-3xcosx)(2)(y+sinx)(1) }{ { (y+sinx) }^{ 3 } } \)
= \(\frac { (y+{ sinx) }-[3y+3sinx-6y-6sinx+6xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ yx }=\frac { -3y-3sinx+6xcosx }{ { (y+sinx) }^{ 3 } } \) ......(1)
\({ f }_{ y }={ [3x[-1][y+sinx] }^{ -2 }\)
\(=\frac { -3x }{ (y+sin{ x) }^{ 2 } } \)
\(\therefore { f }_{ xy }=-3\left[ \frac { ({ y+sinx) }^{ 2 }(1)-x(2)(y+sinx)(cosx) }{ { (y+sinx) }^{ 4 } } \right] \)
\({ f }_{ xy }=\frac { -3(y+sinx)[y+sinx-2xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ xy }=\frac { -3(y+sinx-2xcosx) }{ ({ y+sinx })^{ 3 } } \) ...(2)
From (1) and (3)
f_xy = f^yx