All students will act as the teacher. Each of you will comment on the strategy
that you will use to teach someone about differentiation of polynomials, logs and ln.
All the common errors you have observed can now be avoided, how?
What is your strategy in your teaching to avoid these errors?
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eg:
ReplyDeletey = 1/2X squared.
my approach in this question firstly is to make it look even simplier by separating the 1/2 from the X. and from the rule of differentiation
y = AX power n
dy/dx = AnX power n-1
the above question cana now bw solved as
y = 1/2 X suared
dy/dy = (1/2)(2) X to the power 2-1
(the brackets indicate multiplication)
= 1 X to the power 1
by separating the 1/2 from the X in the above question prevents students from making simple errors
ReplyDeleteSomeone can now differentiate the following question for me please.
ReplyDeletey = 1/3X squard + 1/6X
y= 1/3X cube/3 + 1/6Xsquared/2
ReplyDeletepoopsie, i believe your answer to the question
ReplyDeletey = 1/3X squard + 1/6X
is not correct, you didn't differentiate, i believe u was trying to do integration, but if it was a integration question it would also be wrong, no hard feeling
the answer is
y= 2/3x + 1/6x
but im a little unsure about that answer...
well hellscream i believe your answer is wrong as well because when differentiating a coefficient with jus an x term the differential is the coefficeitn itself.. you dont put back the x.....
ReplyDeletey=1/3x^2+1/6x
dy/dx= 2/3x+1/6
wel firstly i will start on the base to refresh there memories on things they suppose to know about the topic then il give them all the simple concepts or rules, like formulas, i will use simple examples at first and then gradually move on to the the more complecated one, and then finally i will mention to them the hardest thing they can expect in the topic and do allot of examples to make them confident, because as a student i think alot of teachers give students easy examples in class and bring complecated ones for exams.
ReplyDeleteWhat are some of the basic rules when doing log and differentiation? lets see who studies lol. eg. 4^0=1, post some more to refresh others minds plz.
ReplyDeletewell I think craziekenny is also wrong
ReplyDeletethe solution to the question is:
y= (1/3x^2) + (1/6x)
when you bring all the downstairs to upstairs you will have:
y = ((1/3)x^-2)+ (1/6)x^-1
dy/dx = [((1x-2)/3)x^-2-1] + (1x-1/6)x^-1-1)
where dy/dx = ((-2/3)x^-3) - (1/6)x^-2
For differentiation of polynomials we should strive to make all terms (terms are separated by a + or – sign) into the form y=ax^n, where the differential (dy/dx) is simply term (a) multiplied by the power, (n) multiplied by (x) to the power of n-1. Therefore: dy/dx=anx^n-1
ReplyDeleteWhen given a problem, we must first evaluate the following aspects:
1. Look for knotted terms (brackets which are to be worked first or expanded if necessary. E.g y= 8(x2 – 6x) = 8x2 – 48x
The brackets must be expanded first to give terms only
2.Make sure everyone is upstairs, that is, if dealing with fractions ensure that everyone is on the same line. E.g y= 4/x^3 = 4x^-3 y = 4/5x^6 = (4/5)x^-6
3.Remove any shed (square root).E.g. √(x)= x^1/2
3√(x) = x^1/3
4.Look out for any police, i.e. be careful when evaluating the negative signs
well said guniron.
ReplyDeletei eh understand one thing bout differenciation
ReplyDeletewhat about differntiating for sin x, cos x and all that can some 1 plz explain this 2 me. and also what about ln.
ReplyDeleteto differentiate trigonometry functions is very easy 'animechic', all you have to do is remember the basics and you will find that it is easy to do, just remember:
ReplyDeletey = cos x
dy/dx = -sin x
y = sin x
dy/dx = cos x
so from this we can now do problems such as
y = 10 cos q
dy/dx = - 10 sin q
hey but can someone refresh my memory about ln plzz?
ReplyDeletecan anyone help me with this queation>?
ReplyDeleteT is directly proportional d^3
anyone can help i dont understand.plz?????
To mysticwings*, the differentiation of ln is 1/x. Eg.
ReplyDeleteln(3x + 4)
dy/dx = 1/(3x + 4)
Its the easiest..
To small man:
ReplyDeleteTwo quantities, A and B, are in direct proportion if by whatever factor A changes, B changes by the same factor.
e.g.
Case A:
Suppose that you buy 4 marbles. You would pay $2.00.
Case B:
Suppose that you buy 8 marbles. You would pay $4.00.
So, changing the number of marbles that you buy will change the amount of money that you pay.
They increase in a direct proportion.
In mathematical terms:
A is directly proportional to B. To get rid of the directly proportional sign(which i cant find!!, sorry) we replace it by an equal sign and a constant. So it becomes:
A = KB
Then we can find for K.
Concerning your question, T is directly proportional d^3:
It should be:
T = Kd^3
where K is the constant. Then you can differentiate or solve for K, or do whatever you are required to do.
Well small man when a quwstion says
ReplyDeleteT is directly proportional d^3
it is represented by T=Kd^3 where K is a constant. i hope is dat what you did not understand.