Limit for Multivariable Functions
This is just an extension of definition of limit from function of one variable.
The concept can also be extended to more variables. We will only use up to 3 variables.
All Limit properties of one variable functions apply here too.
Continuous Multivariable Functions
Partial Derivatives
• We can extend the notion of differentiation to functions of more than
one variables.
Geometrically, the partial derivative 𝑓𝑥
captures how fast the function changing
in x-direction, and 𝑓𝑦 captures how fast f
is changing in y-direction.
Example
• Find partial derivatives of 𝑓 𝑥, 𝑦 = 𝑥 3 𝑦 2 + 𝑒 𝑥
Example
• Find partial derivatives of 𝑓 𝑥, 𝑦, 𝑧 = 𝑥 2 + 𝑦 2 + 𝑧 2
Higher Order Differetiation
Example
• Find second-order partial derivatives of 𝑓 𝑥, 𝑦 =
𝑦 sin 𝑥 2 .
Order of Differentiation
Therefore
Gradient
• The partail derivatives 𝑓𝑥 and 𝑓𝑦 measures how much f is changing in the direction of x-axis and yaxis, respectively. However, we may also want to know the rate of change in other directions
Example
• Find gradient of 𝑓 𝑥, 𝑦 = ln(𝑥 2 + 𝑦 2 )
Directional Derivative
Example: Find the rate of change of the function 𝑓 𝑥, 𝑦, 𝑧 = 𝑒 𝑥𝑦𝑧 at the point 𝑥, 𝑦, 𝑧 = 1,1,1 in the direction
of vector 𝐰 = −2,1,2
Gradient (what does it measure?)
We will have a lot
more to say about
gradient later.
The Chain Rule
• Recall that chain rule for functions of one variable where g is a function y and y is
𝑑𝑔 𝑑𝑔 𝑑𝑦
a function of x states that = .
𝑑𝑥
𝑑𝑦 𝑑𝑥
• The version of chain rule for functions of more than one variable have a similar
form, but involve more terms.
• The simplest method for generating the chain rule for any particular set of
dependencies of variables is to draw “tree diagram”.
Chain Rule
For example, the tree diagram representing the dependencies for 𝑓 𝑥, 𝑦, 𝑧
where x, y, and z depend on t is
Example
Example
• Find partial derivatives of 𝑓 𝑥, 𝑦 = 𝑥 2 − 𝑦 2 where 𝑥 𝑟, 𝜃 = 𝑟 𝑐𝑜𝑠𝜃,
and 𝑦 = 𝑟 𝑠𝑖𝑛𝜃.
More Examples:
Let 𝑧 = 𝑔 𝑥, 𝑦 and 𝑥 = 𝑠 + 𝑡 and 𝑦 = 𝑠𝑡 Show that:
𝜕2𝑧
𝜕2𝑔
𝜕2𝑔
𝜕 2 𝑔 𝜕𝑔
= 2 +𝑥
+𝑦 2+
𝜕𝑠𝜕𝑡 𝜕𝑥
𝜕𝑥𝜕𝑦
𝜕𝑦
𝜕𝑦
𝑥+𝑦
Let 𝑢 𝑥, 𝑦 = 𝑥𝑦𝑓
where 𝑓: ℝ → ℝ is differentiable. For what value of
𝑥𝑦
𝑔 𝑥, 𝑦 will 𝑢 𝑥, 𝑦 satisfy the partial differential equation:
𝑥 2 𝑢𝑥 − 𝑦 2 𝑢𝑦 − 𝑔 𝑥, 𝑦 𝑢 = 0
Suppose 𝑢(𝑥, 𝑦)is given and 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 . Show that
𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0 equivalent to 𝑟 2 𝑢𝑟𝑟 + 𝑟𝑢𝑟 + 𝑢𝜃𝜃 = 0 (we will use this in our
studies later)
The following is taken directly from final exam paper of ECE 4179,
Sem 1, 2022 “Neural Network and Deep Learning”. (Dr Hisham)