Azhar Rohiman – Mathemafrica

Chain Rule.

Definition:

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

Continuity – (Part Two).

Definition:

(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 x 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 Azhar Rohiman| March 31st, 2016|Courses, First year, MAM1000, Undergraduate|4 Comments

Continuity – (Part One).

Definition:

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

NOTE:

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

About Azhar Rohiman

Chain Rule.

Continuity – (Part Two).

Continuity – (Part One).