Good afternoon all, in a bid to contribute my own quota to "our" admission, I will be treating some MATHEMATICS topics/questions [I am not promising to be online 24/7, but when tides favour the ocean].
Some of the topics "we" will be considering are, in no particular order:
1. Calculus.
2. Trigonometry.
3. Number base.
4. Co-ordinate geometry.
5. Polynomial.
6. Statistics.
7. Matrices.
8. Algebra.
9. Probability.
10. E.t.c.
For any questions/suggestions/observations/e.t.c, please do make it before the first class [23/03/2015].
Note
1. There is no guaranty any of the topics will be among the post utme questions, therefore, I am not liable for any feeling/opinion/e.t.c that may arise from this.
2. I will try my best to make the topics/questions easy to understand, but, typing mathematical symbols/expressions in an online forum is not going to be easy.
3. The fruitfulness of this class depends on my schedule, i.e I can't guaranty everyday lectures.
Vote of thanks:
Mr. Dopeyomi, thank you for this inspiration.
Once again, Happy Sunday to you all.
22nd March, 2015.
13:30:59PM.

Good evening all, as per my obligation, today will be "our" first class. "We" will be considering differentiation.
Note 1:
1. dy/dx can also be represented as y'.
2. d2y/dx2can also be represented as y", till infinity.
3. d/dxmeans "we" are differentiating with respect tox, d/dumeans we are differentiating with respect to . . . . . . . . . . . .very good, u.
4. wrt means with respect to.
DIFFERENTIATION.
Differentiation means rate of change.
General formula:
if y = axn,
a is the co-efficient of x.
n is the power of x.
Therefore, dy/dx = power "times" co-effecient of x "times" the power of x "minus" 1.
Mathematically,
y = axn,
dy/dx = naxn - 1.
-----------------------------
Example 1. Find y' [it also means dy/dx] if y = 2x.
Solution:
the co-efficient of x is 2.
The power of x is 1.
Therefore,
y = 2x
dy/dx = (1)(2)x1-1
dy/dx = 2x0
dy/dx = 2(1)
because x0is 1.
dy/dx = 2.
-----------------------------.
Example 2: find y' if y = 3x4.
Solution:
y = 3x4
y = 12x3
-----------------------------.
Example 3: find y' if y = x
.
.
.
.
.
.
.
y' = 1.
Very good, because,
y = x
the co-efficient of x is 1.
The power of x is 1.
Therefore,
y = x
y' = (1)(1)x1-1
y' = 1x0
y' = 1(1)
y' = 1.
-----------------------------.
Example 4: find y' if y = 2.
Solution:
y = 2,
the question can also be changed into
y = 2(1)
it can be further changed into
y = 2(x0)
since,
y = 2x0
y' = (0)(2)x0-1
y' = 0x-1
y' = 0.
Note 2: The derivative of a constant is 0.
i.e
d/dx (7) = 0
d/dx (10) = 0, but,
d/dx (7x) = 7
d/dx (10x) = 10.
-----------------------------.
Summary:
1. Differentiation means rate of change.
2. If y = axn, then dy/dx = naxn-1.
3. The differentiation of constant values is 0. E.g,
y = 6,
dy/dx = 0,
y = 1000000000
dy/dx = 0.
-----------------------------.
Assignment: find dy/dx if y =
1. 20x
2. -50x3
3. -25x-2
4. 30/x2, [Hint: 1/x2is x-2].
Any question?
Good night class.
23rd March, 2015.
21:35:45

Good morning to you all, how was the night?
Yesterday, we were considering differentiation, we were able to conclude that:
1. Differentiation means rate of change.
2. If y = axn, then dy/dx = naxn-1.
3. The differentiation of a constant is zero.
Today, we will be considering another aspect of differentiation.
DIFFERENTIATION OF POLYNOMIALS.
To differentiate a polynomial, we differentiate each term in turn.
General form:
if y = axb+ cxd+ exf+ . . .,
then dy/dx = baxb-1+ dcxd-1+ fexf-1+ . . ..
. . . . . . . . . . . . . . . . . . . . . . . .
Example 1: if y = x4+ 5x3- 4x2+ 7x - 2.
Solution:
dy/dx = (4)(1)x4-1+ (5)(3)x3-1- (4)(2)x2-1+ (7)(1)x1-1- 0.
dy/dx = 4x3+ 15x2- 8x + 7.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 2: if y = 2x5+ 4x4- x3+ 3x2- 5x + 7, find an expression for dy/dx and the value of dx/dx at x = 2.
Solution:
So, first of all,
dy/dx =
.
.
.
.
.
.
.
Very good, because, if
y = 2x5+ 4x4- x3+ 3x2- 5x + 7.
Then,
y' = 10x4+ 16x3- 3x2+ 6x - 5.
Substituting x = 2, we have,
y' = 10(2)4+ 16(2)3- 3(2)2+ 6(2) - 5.
Therefore,
y' = 283.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 3:
if y = 2x2- 5x + 3, then
dy/dx = 4x - 5.
This double statement can be written as a single statement by putting 2x2- 5x + 3 in place of y in dy/dx. i.e
d/dx (2x2- 5x + 3) = 4x - 5.
. . . . . . . . . . . . . . . . . . . . . . . .
Example 4:
In the same way,
d/dx (4x3- 7x2+ 2x + 3) =
Solution:
12x2- 14x + 2.
Note: Either of the two methods is acceptable: it is just a case of which is the more convenient in any situation.
. . . . . . . . . . . . . . . . . . . . . . . .
Summary:
1. To differentiate a polynomial, we differentiate each term in turn.
2. if y = axb+ cxd+ exf+ . . .,
then dy/dx = baxb-1+ dcxd-1+ fexf-1+ . . ..
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
1. No 39, 1994.
If y = 3t3+ 2t2- 7t + 3, find dy/dt at t = -1.
(a) -1 (b) 1 (c) -2 (d) 2.
2. No 19, 2002.
If y = x2- 1/x, find dy/dx
(a) 2x + 1/x2(b) 2x + x2(c) 2x - 1/x2(d) 2x - x2
3. No 36, 2008.
Find the derivative of y = x7- x5
(a) x (x2 - 1) (b) 3x (x[sup]2- 1) (c) 3x2- 1 (d) 7x6- 5x4.
. . . . . . . . . . . . . . . . . . . . . . . .
Things to note before the next class:
1. There is difference between
(a) sinx - cosx
(b) sinx + cosx
(c) sinx cosx,
the difference is, (a) &(b) are polynomials because they are connected by either + or - while (c) is connected by multiplication. Another method is used to solve c.
2. If y = sinx, dy/dx = cosx.
3. If y = cosx, dy/dx = -sinx.
Have a splendid day.
Any question?
24th March, 2015.
07:07:33

Good evening all, how is your day fairing?
In our last class, we were considering differentiation of polynomials. We were able to achieve that
1. To differentiate a polynomial, we differentiate each term in turn.
2. If y = axb+ cxd+ . . ., then, dy/dx = baxb-1+ dcxd-1+ . . ..
Now, we will be considering a different aspect:
HIGHER DERIVATIVES.
If y = 2x4- 5x3+ 3x2- 2x + 4, then, by the previous method:
dy/dx = 8x3- 15x2+ 6x - 2.
This expression for dy/dx is itself a polynomial in powers of x and can be differentiated in the same way as before, i.e. we can find the derivative of dy/dx.
d/dx [dy/dx] is written d2y/dx2and is the second derivative of y wrt x (spoken as "dee two y by dee x squared".
So, in this example, we have:
y = 2x4-5 x3+ 3x2- 2x + 4.
dy/dx or y' = 8x3- 15x2+ 6x - 2.
d2y/dx2or y" = 24x2- 30x + 6.
We could, if necessary, find the third derivative of y in the same way:
d3y/dx3or y"' = . . . . . . . . . .
.
.
.
Very good,
y"' = 48x - 30
. . . . . . . . . . . . . . . . . . . . . . . . Summary:
if y = axb+ cxd+ exf+ gxh+ . . .
Then,
dy/dx = baxb-1+ dcxd-1+ fexf-1+ hgxh-1+ . . .
Also,
y" = (b-1)baxb-1-1+ (d-1)dcxd-1-1+ (f-1)fexf-1-1+ (h-1)hgxh-1-1+ . . .
. . . . . . . . . . . . . . . . . . . . . . . .
Things to note before next class:
1. Y = sinx
dy/dy = cosx
2. Y = cosx
dy/dx = -sinx
3. Y = tanx
dy/dx = sec2x.
. . . . . . . . . . . . . . . . . . . . . . . .
Assignment:
Pick up JAMB mathematics past question and solve anything "solvable" in differentiation.
Any question?.
Have a blissful night rest.
24th March, 2015.
19:22:36

GOOOOOOOOOD WORK

this is so cool man!!I'm loving Maths.

Nice job, man.

shadowgwalker:
this is so cool man!!I'm loving Maths.

Avast:
GOOOOOOOOOD WORK

Girltee1:
Nice job, man.

check the UI 2015/2016 for current update or better still, you can follow me to be updated.

I need to continue this, not for JAMBITES but to stimulate interest and reduce anxiety.