Question

Tutorial
© If u = x²y-xsin (xy) find du/dx and du/dy​

Answer 1

= k T V − 1 ,
∴∂p = −kTV−2=−kT. ∂V V2

Section 1: Partial Differentiation (Introduction) 5
The symbol ∂ is used whenever a function with more than one variable is being differentiated but the techniques of partial differentiation are exactly the same as for (ordinary) differentiation.
Example 2 Find
∂z and ∂z for the function z = x2y3. ∂x ∂y
Solution
and ∂z ∂y
∴ ∂z ∂x
= 2xy3 , = x23y2 ,
z = x2y3
For the first part y3 is treated as a constant and the derivative of x2 with respect to x is 2x.
For the second part x2 is treated as a constant and the derivative of y3 with respect to y is 3y2.
(Click on the green letters for solutions.)
(a) z = x2y4, (b) z = (x4 +x2)y3, (c) z = y12 sin(x).
= 3x2y2 .
Exercise 1. Find ∂z and ∂z for each of the following functions.
∂x ∂y

Section 2: The Rules of Partial Differentiation 3
1. Partial Differentiation (Introduction)
In the package on introductory differentiation, rates of change of functions were shown to be measured by the derivative. Many applications require functions with more than one variable: the ideal gas law, for example, is
pV = kT
where p is the pressure, V the volume, T the absolute temperature of
the gas, and k is a constant. Rearranging this equation as p = kT
V
shows that p is a function of T and V . If one of the variables, say T,
is kept fixed and V changes, then the derivative of p with respect to V measures the rate of change of pressure with respect to volume. In this case, it is called the partial derivative of p with respect to V and written as
∂p . ∂V

Section 1: Partial Differentiation (Introduction) 4
Example 1 If p = kT , find the partial derivatives of p: V
(a) with respect to T , (b) with respect to V . Solution
(a) This part of the example proceeds as follows: p = kT,
V
∴ ∂p = k, ∂T V
where V is treated as a constant for this calculation. (b) For this part, T is treated as a constant. Thus
p = k