Qn frm email (Additional Mathematics Question / A-Math Questions)
Differentiate ln square root (2x+4)
Hint#1: Bring out Laws of Ln which is somewhat the same as Laws Of Log :)
Hint#2: When you are asked to Diff Ln (whatever) = Diff (whatever) OVER (whatever) :)
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Amaths Differentiation and Integration techniques.
How to actually remember all the different ways for differentiating and Integrating, I always get mixed up
Hi xxx,
Differentiation = Decrease Power
Integration = Increase Power
(It's often sth OVER another)
Example
Diff e^2x = 2e^2x
Integrate e^2x = (e^2x / 2)
Diff (3x+4)^5 = 5(3x+4)^4 (3)
Integrate (3x+4)^5 = (3x+4)^6/(3*6)
It's suggested that you have a "cheat" sheet with Diff on Left Hand Side and Integration on Right Hand Side.
To familiarise more, work on at least 5 qns just on differentiation and integration :)
You can always scan me any qns or type out any should you face any challenges.
Cheers,
alwaysLovely
Ai Ling Ong
Hi, I'm Ai Ling Ong. I enjoy coaching students who have challenges with understanding and scoring in 'O' Level A-Maths and E-Maths. I develop Math strategies, sometimes ridiculous ideas to help students in understanding abstract concepts the fast and memorable way. I write this blog to share with you the stuff I teach in my class, the common mistakes my students made, the 'way' to think, analyze... If you have found this blog post useful, please share it with your friends. I will really appreciate it! :)
thank u v much for the hints on integration .
i always look forward to such challenging problems.
binomial expansion problem, i do the problems always the same method----term independent of x .
keep guiding us.
thanks love
Meera,
Thanks for your kind words.
I did a search on Binomial Expansion and found a post which addresses the concept of Term Independent of x!
I hope you find this post useful too.
Click on the link to read more:
http://www.singaporeolevelmaths.com/2009/01/30/a-math-binomial-expansion-finding-term-independent-of-x-by-a-shortcut-method/
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Thank you for your wonderful comment.
I'm glad that I can help.
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