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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:

differentiation-qn

Very often, I notice students will jump into Quotient rule whenever a fraction is given. Just like this student here:

differentiation-qn-quotient-rule

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:

differentiation-qn-simplified

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?

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Jan/08

31

Differentiation Basic Techniques

calculus.jpg

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.

  1. Chain Rule
  2. Product Rule
  3. 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 :

  1. Leave the constant in front
  2. 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

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About Ai Ling

Hi, my name is Ai Ling Ong and I'm a professional Math Trainer specializing in GCE 'O' level Mathematics (Additional & Elementary). More than 10 years of teaching experience, I have coached hundreds of students more than 80 schools in Singapore. Read more about me here
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