In the tutorial, we will learn the partial derivatives for multiple variable functions.

Assume is a two variable function, then we use the notations to describe the partial derivatives of

denotes the partial derivative of f under the variable x.

denotes the partial derivative of f under the variable y.

Similarly, we can also define the second derivative of

,

.

Assume is a unit vector, i.e. its length is 1. If is at the point p, then we can define the **directional derivative** of at point p as

**Theorem 1. Geometric mean is not larger than Arithmetic mean.**

For n positive real numbers ,

“=” if and only if

**Theorem 2. Cauchy’s Inequality.**

For 2n real numbers ,

“=” if and only if

**Proof.**

**Method (i). **Construct a non-negative function f(x) with respect to variable x

Consider the equation f(x)=0, there are only two possibilities: one is the equation f(x)=0 has only one root, the other one is the function has no real roots. Therefore,

Hence,

Moreover, if “=”, then f(x)=0 has only one root , i.e. for all , . That means

By the way, the solution of is and

**Method (ii). **Since , we know

for all

Assume , for all ,

Take the summation at the both sides,

**Question 1.** Assume is a function and . satisfies the partial differential equation

Prove

(1) is a function of y.

(2) is a function of x.

(3) The solution of has the form for some function and .

Proof.

(1) **Method (i) **Make use of derivative.

First, we know . Second, take the partial derivative of the function with respect to the variable x. That means,

from the partial differential equation. Therefore, the function is independent of the variable x. i.e. the function is a function of variable y.

**Method (ii) **Make use of integration.

Since , , then we take the integration of x at the both sides,

, the left hand side is , the right hand side is for some function That means, . and is a function of .

(2) is similar to (1).

(3) From part (1), we know is a function of y. Assume . Take the integration of y at the both sides, we have

for some function for some functions and

**Question 2. **Assume and are non-negative. Find the maximum value of .

**Solution.**

**Method (i)**. Langrange’s Method.

Take three partial derivatives of g,

Solve these three equations, we get and , therefore the maximum value is taken at and

**Method (ii)**. Change to one variable function.

Since L+K=150, we can define the one variable function

The derivative of

The critical point is The maximal value of g(L) is taken at

**Method (iii)**. Mathematical Olympic Method.

Use the fact that the geometric mean is not larger than the arithmetic mean.

.

The maximum value is taken at i.e.

**Question 3.** Assume and are non-negative variables. . Find the maximum value of

**Solution.**

**Method (i).** Langrange’s Method

Take four partial derivatives of the critical point is taken at i.e. the maximum value of f(x,y,z) is taken at

**Method (ii)** Math Olympic Method

The maximum value is taken at , i.e.

**Question 4. ** 2012 Exam MA1505 Semester 1, Question 3(a)

Assume has continuous partial derivatives of all orders, if

Find the value of the constant

**Solution.**

**Method (i) **Use derivatives.

Since has continuous partial derivative of all orders,

Since and

we have

This implies i.e.

**Method (ii). **Use integration.

Comparing them, we know and where and are constants.

Therefore