Let f(x) = tan x, x_o= 45, dx = 1
f(x_o) = f(45) = tan 45 = 1
f'(x) = sec² x dx
f'(x_o) = f'(45) = sec² 45 (1)
\((\sqrt{2})^2\) =2 (0.01745)
= 0.03490
∴ tan 46° = f(x_o) +f'(x_o) dx
=- 1 + 0.03490 = 1.03490

Let x_o = 3 and dx = 0.02
f(x_o) = f(3) = 3 (3²) + 5 (3) + 3
= 27 + 15 + 3 = 45
f'(x) = 6x + 5
f'(x) = f'(3) = 6 (3) + 5 = 23
∴ f(3. 02) = f(x_o) +f'(x_o) dx
= 45 + 23 (.02)
= 45 + 0.46 = 45.46

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).

Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g 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).

Given absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

\({ h }({ x }_{ o })=\frac { x }{ 1+1 } =\frac { 1 }{ 2 } \)
\({ h }^{ ' }(x)=\frac { (x+1)(1)-x(1) }{ { (x+1) }^{ 2 } } \)
\(\frac { x+1-x }{ { (x+1) }^{ 2 } } =\frac { 1 }{ ({ x+1) }^{ 2 } } \)
\({ h }^{ ' }({ x }_{ o })=\frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ 4 } \)
∴ 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) = \(\frac { 9-4x }{ 5 } \)

Given r² = x² +y² + z²
log r² = log (x² + y² + z²)
⇒ 2 log r = log (x² + y² + z²)
∴ 2V = log (x² + y² + z²) [∵ V = log r ]
⇒ V = \(\frac12\) log (x² + y² + z²)
\(\frac { \partial V }{ \partial x } =\frac { 1 }{ 2 } \frac { 2x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } =\frac { x }{ { x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } } \)
\(\frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } =\frac { ({ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 })(1)-x(2x) }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-{ x }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
III^ty \(\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ y }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
∴ \(\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { { (x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }) }^{ 2 } } \)
\(\therefore \frac { { \partial }^{ 2 }V }{ \partial { x }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { y }^{ 2 } } +\frac { { \partial }^{ 2 }V }{ \partial { z }^{ 2 } } =\frac { { y }^{ 2 }+{ z }^{ 2 }-{ x }^{ 2 }+{ z }^{ 2 }+{ x }^{ 2 }-{ y }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 }-{ z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } \)
= \(\frac { { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 }) }^{ 2 } } =\frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 }{ +z }^{ 2 } } \)
= \(\frac { 1 }{ { r }^{ 2 } } \)
Hence proved.

We need to show that u satisfies the Laplace’s equation in R². Observe that u_x(x, y) = e^-2y(-2)sin(2x) and hence u_xx (x, y) = e^-2y(-2)(2) cos(2x).
Similarly, u_y( x y) = e^-2y (-2)cos(2x) and u_yy (x, y) = (-2)(-2)e^-2ycos(2x)
Thus, u_xx + u_yy = -4e^-2y cos(2x) + 4e^-2y cos(2x) = 0.

Given f(x,y) = x² + y³ + 2xy²
f_x = 3x² + 2y²
f_xx = 6x
f_y = 0+ 3y² + 4xy
= 3y² + 4xy
f_yy = 6y + 4x
f_xy = 4y
f_yx = 4y

Given f(x, y) = 2x³ - 11x²y + 3y³
f(tx, ty) = 2t³ x3 - 11 t² x²ty + 3t³y³
= t³(2x³ - 11x²y + 3y³)
= t³. f(x,y)
∴ f (x, y) is a homogeneous function of degree 3.
∴ By Euler's theorem,
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =3f\)