TAG | chain rule
28
A-Math: Differentiation Shortcut Lies In Pausing & Simplifying
2 Comments | Posted by alwaysLovely in A-Math
Differentiation is a big thing in fact major chapter for all Secondary 4 ‘O’ level students.
Read all about the basics Differentiation techniques here. (Examples included) I would like to share one question from my A-Math Ultimate Leap Programme (weekly coaching class) which has 2 different approaches to solve it.
Example:
Very often, I notice students will jump into Quotient rule whenever a fraction is given. Just like this student here:
May I suggest that you pause for 3 seconds to think about the approach. Ask yourself ‘Is there anything I can simplify?’
Here’s another student who pauses:
Notice this student spends his time simplifying before applying chain rule in differentiation.
I hope you enjoy this example. Both students are correct in their answers, which one do you prefer more? A or B?
31
Differentiation Basic Techniques
3 Comments | Posted by alwaysLovely in A-Math, Learning Tools

The big thing for Sec 4 A-Math ( Oh yar, since this blog has quite a number of visitors from other countries, I think it is important for me to mention that Secondary 3 – 4 is known as Grade 9 – 10 in other countries) is CALCULUS.
Well, you will hardly find this word in your A-Math textbook but you see a bulk of the chapters dedicated to Differentiation & Integration. These 2 topics arelike freezing and melting processes.
Why? Because they are simply opposite of each other !
I am going to talk about the techniques of differentiation.
There are 3 main types for Sec 4 level ; you ought to learn these techniques real well and know when to apply each one of them as application problems follow after the basics.
- Chain Rule
- Product Rule
- Quotient Rule reserved for fractions. * But some fractions can skip this rule
My personal favourite is No. 2 - Product Rule. Well, let’s see the technique in action.
Differentiate [tex]y=3 (x^2 + 5)^4[/tex] with respect to x:
Now to do this, you can apply Product Rule – Differentiate Copy + Differentiate Copy
so [tex]\frac {dy}{dx}=0 (x^2 + 5)^4+ 4(x^2 + 5)^3(2x)(3)
=24x(x^2 + 5)^3[/tex]
Now from this example, we notice some patterns, if you have a constant ( a fixed number) in front, you can simply focused on differentiating the portion with x involved. For example in this case, Focus on Differentiating [tex](x^2 + 5)^4[/tex] so now, we don’t even have to use product rule :
- Leave the constant in front
- Differentiate the portion with x involved by Power Front – Power Down by 1-Differentiate within also known as your Chain Rule. BINGO!
[tex]\frac {dy}{dx}=(3) 4(x^2 + 5)^3(2x)=24x(x^2 + 5)^3[/tex]
By realizing this pattern, it will save you some time and less pen ink as well.
Certainly hope it is useful.
:-)
alwaysLovely
Photo by just another paul












