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

Note that g is differentiable and g'(x) = 2x + cos x
Thus dg = (2x + cos x)dx.

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)
= \(\frac { dF }{ dt } \)

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.

First we calculate  \(\frac { { \partial }w }{ { \partial x } } (x,y)=\frac { { \partial }(xy) }{ { \partial x } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial x } \)
This gives  \(\frac { { \partial }^{ }w }{ { \partial x } } \) (x, y) = y + 0 and hence \(\frac { { \partial }^{ 2 }w }{ \partial y\partial x } \) (x, y) = 1 On the other hand,
\(\frac { { \partial }w }{ { \partial y } } (x,y)=\frac { { \partial }(xy) }{ { \partial y } } +\frac { \partial \left( \frac { { e }^{ y } }{ { y }^{ 2 }+1 } \right) }{ \partial y } \)
\(=x+\frac { \left( { y }^{ 2 }+1 \right) { e }^{ y }-{ e }^{ y }2y }{ \left( { y }^{ 2 }+1 \right) } \)
Hence, \(\frac { { \partial }^{ 2 }w }{ { \partial x\partial y } } \) (x, y) = 1

First we shall calculate \(\frac { \partial F }{ \partial x } \)(x, y) then we evaluate it at (−1, 3) As we have already observed we find the derivative with respect to x holding y as a constant. That is,
\(\frac { \partial f }{ \partial x } (x,y)=\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial x } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial x } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial x } +\frac { \partial (7) }{ \partial x } \)
= 3x² y + y² +0
= 3x² y + y² .
so, \(\frac { \partial F }{ \partial x } \) ( -1, 3) = 3( -1)² 3 + 3² = 18.
Next similarly we find partial derivative with respect to y.
\(\frac { \partial F }{ \partial y } \) (x, y) = \(\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial y } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial y } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial y } +\frac { \partial (7) }{ \partial y } \)
= x³ + 2yx + 0
= x³ + 2yx.
Hence we have \(\frac { \partial F }{ \partial y } \) (-2, 1) = (-2)³ + 2(1)( -2) = -12.
Note that in the above example \(\frac { \partial F }{ \partial x } \) (x, y) = 3x² y + y² which is again a function of two variables. 
So, we can take the partial derivative of this function with respect to x or y.
For instance, if we take G(x, y) = 3x² y+y² then we find  \(\frac { \partial F }{ \partial x } \) = 6xy. Since G(x, y) = \(\frac { \partial F }{ \partial x } \), we have \(\frac { \partial G }{ \partial x } \)=\(\frac { \partial }{ \partial x } \)\(\left( \frac { \partial f }{ \partial x } \right) \) = 6xy.
We denote this as \(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } \) which is called the second order partial derivative of F with respect to x. 
Also, \(\frac { \partial F }{ \partial y } \) = 3x² + 2y. Since G (x, y) = \(\frac { \partial F }{ \partial x } \) we have  \(\frac { \partial G }{ \partial y } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \) = 3x² + 2y. 
We denote this as \(\frac { { \partial }^{ 2 }F }{ \partial y\partial x } \) which is called the mixed partial derivative of F with respect to x, y. 
Similarly we can also calculate \(\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \right) \) = 3x²+2y.
Also, if we differentiate \(\frac { \partial F }{ \partial y } \) partially with respect to y we obtain \(\frac { { \partial }^{ 2 }F }{ { \partial y }^{ 2 } } \) which is called the second order partial derivatives of F with respect to y.
So for any function F defined on any subset {(x, y) | a < x < b, c < y < d} ⊂ R² we have the following notation
\(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ xx' }\frac { { \partial }^{ 2 }F }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ xy }\)
\(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ yx' }\frac { { \partial }^{ 2 }F }{ { { \partial y }^{ 2 } } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ yy }\)
All the above are called second order partial derivatives of F. 
Similarly we can define higher order partial derivatives.
For example, \(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) and \(\frac { { \partial }^{ 2 }F }{ { \partial x\partial y\partial x } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) 
Next we shall see more examples on partial differentiation.

Note that the function f takes value 1 on the x, y-axes and 0 everywhere else on R². So let us calculate
\(\frac { \partial f }{ \partial x } (0,0)\) = \(\underset { h\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0+h,0 \right) -f(0,0) }{ h } =\underset { h\longrightarrow 0 }{ lim } \frac { 1-1 }{ h } =0;\)
\(\frac { \partial f }{ \partial y } (0,0)\) = \(\underset { k\longrightarrow 0 }{ lim } \) \(\frac { f\left( 0,0+k \right) -f(0,0) }{ k } =\underset { k\longrightarrow 0 }{ lim } \frac { 1-1 }{ k } =0\)
This completes (i).