About Azhar Rohiman

I'm a student at the University of Cape Town, currently studying Computer Science and Mathematics.

Chain Rule.


The chain rule is a method for differentiating a function of a function, or differentiating composite functions.

Consider the expression y = sin(x^2). We notice that this is not a normal sine function. It has an x^2 as argument for the sine function. Therefore, we can consider the x^2 in the sine function as a whole different function. This can be broken into two functions, f(x) and g(x).

If we consider f(x) =  sin(x); g(x) = x^2, we can write y = f(g(x)).

In order to differentiate a composite function, y = f(g(x)), i.e; to find y', we let

u = g(x)  and  y = f(u)

What this implies is that whatever g(x) is, u will be equal to that. Then, the process of differentiating is to find

  • \frac{du}{dx} (as u will be a function of x).
  • \frac{dy}{du}

Finally, we can write,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Back to our example, y = sin(x^2); remember that f(x) = sin(x) and g(x) = x^2. We let

u =  x^2  and  y = f(u) = sin(u)

Therefore,  \frac{du}{dx} = 2x and  \frac{dy}{du} = cos(u)

This leaves us with  \frac{dy}{dx} = cos(u) \cdot 2x. We can simplify the equation by writing

\frac{dy}{dx} = 2xcos(x^2)

Note that the u in the cosine function is replaced with x^2.…

By | April 29th, 2016|Courses, First year, MAM1000, Undergraduate|0 Comments

Continuity – (Part Two).


(i) A function f is said to be continuous from the right at a if

\lim\limits_{x \to a^{+}} f(x) = f(a)


We can see that, as the function approaches a certain x-value from the right, f is defined and

\lim\limits_{x \to a^{+}} f(x) \equiv f(a)

And as the function approaches a certain x-value from the left, f is not defined, i.e;

\lim\limits_{x \to a^{-}} f(x) \neq f(a).

Therefore, we say that the function is continuous from the right at this point, but is discontinuous from the left.

(ii) A function f is said to be continuous from the left at a if

\lim\limits_{x \to a^{-}} f(x) = f(a)


Here, it is clear from the graph that the function is continuous from the left as approaches 3. This is because the function is defined at x = 3 and,

\lim\limits_{x \to 3^{-}} f(x) \equiv f(3)

However, from the right,

\lim\limits_{x \to 3^{+}} f(x) \neq f(3)

So, we can say that the function is continuous from the left, but discontinuous from the right, at the point x = 3.

We may also have a function where it is neither continuous at a point from the left or from the right but is defined elsewhere.…

By | March 31st, 2016|Courses, First year, MAM1000, Undergraduate|4 Comments

Continuity – (Part One).


A function f(x) is continuous at a given point x = a if those three conditions below are met “simultaneously”:

(i) f(a) is defined. (i.e; a is in the domain of f)

(ii) \lim\limits_{x \to a} f(x) exists.

(iii) \lim\limits_{x \to a} f(x) = f(a).


  • If any one of the three conditions is false, then f is discontinuous at a, or it has a discontinuity at a.

Let’s now look at the different cases where f(x) may not be continuous at x = a.

(i) f(a) is defined but \lim\limits_{x \to a} f(x) does not exist.


At a = 0, the function is not continuous despite f(a) is defined (Here, f(a) is equal to -1). This is because the two one-sided limits are not equal and as a consequence, the limit does not exist. This is called a jump discontinuity.

(ii)  \lim\limits_{x \to a} f(x) exists, but f(a) is not defined.


Assume a is the x-value where there is a hole in the graph. We can see that the limit from the right of a and the limit from the left of a are equal.…

By | March 31st, 2016|Courses, First year, MAM1000, Uncategorized, Undergraduate|5 Comments