Lecture 1 – partial derivatives & total differentiation
Partial Differentiation
Partial derivatives – the rate of change of the value of
the function due to the corresponding change in that
independent while all the other independent variables are
held constant.
∂Y
∂X
instead of
dY
dX
Z = f(X, Y W)
The partial derivative of z with respect to X can be
denoted in the following different forms:
∂Z
∂X
or
∂
∂X
(𝑧) or
∂f
∂X
Y =f(X1, X2 ….X3)
∂Y
∂X3
or
∂f
∂X3
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Lecture 1 – partial derivatives & total differentiation
Y = f(x1, x2) = x13 + x12x22 + x23
∂Y
∂X1
∂Y
∂X2
= 3x12 + 2x1x22
= 2x12x2 + 3x22
Y = f(u,v) = u2 + uv + v2
∂Y
∂U
∂Y
∂V
= 2u + v
= u + 2v
Total Derivatives
Z = f(x, y) and y = g(x)
Z is expressed as a function of the two variables X. In this
system, X can be referred to as the ultimate variable, it is
the variable with respect to which differentiation will be
carried out.
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Lecture 1 – partial derivatives & total differentiation
The ultimate variable X affects Z from two angles:
1. X affects Z directly through the function f
2. X also affects Z indirectly through the function G via
Y.
To find the effect of a change in X on Z, we apply total
differentiation which will generate the direct as well as
the indirect effects of X on Z.
The total derivative of Z w.r.t. X is a measure of the total
effect (the direct and indirect effects) of the change in X
on the variable Z. The process of computing the total
derivative is referred to as total differentiation.
Z = f(x, y) and y = g(x)
Step 1
Find the total differential of Z.
dz = fxdx + fydy ……………………….. (1)
Step 2
Divide equation 1 all thru by dx and simply.
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Lecture 1 – partial derivatives & total differentiation
dz
dx
dz
dx
dz
dx
= fx
dx
dx
+ fy
= fx + fy
=
∂z
∂x
+
dy
dx
dy
………………………. (2)
………………………… (3)
dx
∂z dy
∂y dx
(i)
Direct effect of X on Z through the function z
f(x, y)
(ii)
Indirect effect of X on Z via Y is:
fy
dy
dx
=
∂z dy
∂y dx
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Lecture 1 – partial derivatives & total differentiation
Exercise
Find the total derivative of Z w.r.t. X for the following
functions
(a) Z = f(x, y) = 5x3 + 3y2
(b) Z = f(x, y) = x4 + 2y3
y = 4x2
y = x2 + 2x + 7
Suggested Solutions
(a) Z = f(x, y) = 5x3 + 3y2
y = 4x2
 Find the total differential of Z.
dz = fxdx + fydy =
∂z
∂x
dx +
= 15x2dx + 6ydy
 Divide all thru by dx.
dz
dx
2 dx
= 15x
dx
+ 6y
dy
dx
= 15x2 + (6y)(8x)
= 15x2 + 48xy
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∂z
∂x
dy
Lecture 1 – partial derivatives & total differentiation
(b) Z = f(x, y) = x4 + 2y3
y = x2 + 2x + 7
We can also use the formula:
∂z
∂x
dz
dx
3
= 4x ;
=
∂z
∂x
+
+
∂z dy
∂y dx
∂z dy
∂y dx
dz
dx
=
∂z
∂x
+
∂z dy
∂y dx
= (6y2)(2x + 2)
= 4x3 + (6y2)(2x + 2)
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Lecture 1 – partial derivatives & total differentiation
Composite Functions
Z = f(x, y)
X =g(u), and Y =h(u)
 This function has two interior functions, X =g(u) and
Y =h(u)
 The ultimate variable is U, thus, total differentiation is
conducted with respect to U.
Step 1
Totally differentiate Z
dz = fxdx + fydy ……………………….. (1)
Step 2
Divide equation 1 all thru by du.
dz
du
= fx
dx
du
+ fy
dy
du
………………………. (2)
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Lecture 1 – partial derivatives & total differentiation
dz
du
=
∂z dx
∂x du
+
∂z dy
∂y du
……………………… (3)
The total derivative of Z with respect to U is
given by equation 3. The two components of
the total derivative are:
(i)
𝑓𝑥
(ii) 𝑓𝑦
dz
du
dz
du
=
=
∂z dx
∂x du
∂z dy
∂y du
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Lecture 1 – partial derivatives & total differentiation
Find the total derivatives of the following functions:
(a) Z = 6x2 + 4y2
X = 3u
Y = 2u
(b) Z = x4 -3y3
X = u2
Y = u3 – 2u
Suggested Solutions:
(a)
dz
du
∂z
∂x
∂z
∂y
dz
du
=
=
∂z dx
∂x du
+
∂z dy
∂y du
= 12x
∂x du
+
du
dy
= 8y
∂z dx
dx
du
∂z dy
∂y du
=3
=2
= 12x(3) + 8y(2)
= 36x +16y
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Lecture 1 – partial derivatives & total differentiation
Second-order-derivatives – measures the response of the
first derivative to changes in the independent variable of
the function.
Z = f(x, y)
F.OPD
SOPD
∂z
∂x
;
∂z
∂y
fxx(x, y) or fxx or
fyy(x, y) or fyy or
𝜕2 z
∂𝑥 2
𝜕2 z
∂𝑦 2
Cross partial derivative of fy with respect to x.
𝜕2 z
∂x ∂y
or
∂ ∂z
∂x ∂y
or
𝜕2 f
∂x ∂y
or fxy
Similarly we can obtain the cross partial derivative of fx
with respect to Y.
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Lecture 1 – partial derivatives & total differentiation
Z =(x, y) = 5x3y2
Find the 1st, 2nd orders and cross partial derivatives:
(i)
FOPD
∂z
∂x
(ii)
𝜕2 z
∂𝑥
∂z
= fx = 15x2y2
= fy = 10x3y
∂y
SOPD
2 = fxx = 30xy
𝜕2 z
2
∂𝑦
3
=
f
=
10x
yy
2
Cross partial derivatives: differentiating with respect to y
and x: fx and fy
𝜕2 z
∂x ∂y
𝜕2 z
2
= fxy = 30x y
∂y ∂x
= fyx = 30x2y
Young’s theorem – if all the partial derivatives of the
function do exist and if the partial derivatives are smooth,
continuous, differentiable and have smooth, and
continuous derivatives, then the cross-partial derivatives
derived from the function will be equal.
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Lecture 1 – partial derivatives & total differentiation
Implicit Function Differentiation
Implicit functions – those functions in which both the
dependent and the independent variables appear on the
same side of the equal sign. Dependent and independent
variables are not clearly separated.
2x2y4 – xy2.7 + x3 + 3y2 = 0
4x12x3y2 – 3x1y2 + x1x3y2 = 18
F(y, x1, x2, x3) = C, implicit function presented in the
general form dependent variable y and independent
variables, x1, x2 and x3.
Y = x12 + x22 ------------3x12
Y = f(x1, x2 ----------- xn)
F(X, y) = 50
function
explicit function
implicit function: differentiate this
1. Totally differentiate the function
fxdx + fydy = 0
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Lecture 1 – partial derivatives & total differentiation
2. Rearrange the terms:
fydy = -fxdx
divide both side sides by dx provided that fy = 0.
dy
dx
=−
fx
formula for differentiating the
fy
implicit functions states that the derivative
the negative of the ratio
dy
dx
is equal to
fx
fy
Q(K, L) = A, differentiate the production function
Solution
 Totally differentiate
QKdK + QLdL = 0
 Rearrange and obtain
dK
dL
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Lecture 1 – partial derivatives & total differentiation
QKdK = -QLdL
dK
dL
=−
QK
QL
Differentiate the following implicit function:
4x3 + 3y2 – 13 = 0
dy
dx
= 12x2dx + 6ydy = 0
6ydy = -12x2dx
dy
dx
=-
12𝑥 2
6y
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