.

vickvan
If y =3sin (-4x) find dy/dx ............................... let u=-4x therefore y=3sinu
du/dx = -4
dy/du = 3 cos u
from function of a function formula
dy/dx = dy/du × du/dx
in summary, dy/dx = 3cosu × -4
dy/dx = -12cosu
remember u=-4x
therefore dy/dx = -12cos (-4x)

vickvan:
If y= 3sin (-4x), dy/dx is

I think you'll use product and chain rule.

Product rule
V.du/dx + U.dv/dx

u = 3, v =sin(-4x)

du/dx = 0, dv/dx =? Using chain rule: let a = -4x

:. V = sina

chain rule states, dv/dx = dv/da * da/dx

dv/da = cosa, da/dx = -4

:. dv/dx = cosa * -4 = -4cosa

back to product rule which states:

dy/dx = V.du/dx + U.dv/dx

v = sin(-4x), u =3, du/dx = 0, dv/dx = -4cosa

therefore, dy/dx = sin(-4x) * 0 + 3 * -4cosa

-12cosa. Recall that a = -4x

dy/dx = -12cos(-4x)...


Abeg who knows a quicker method?

Geofavor:


I think you'll use product and chain rule.

Product rule
V.du/dx + U.dv/dx

u = 3, v =sin(-4x)

du/dx = 0, dv/dx =? Using chain rule: let a = -4x

:. V = sina

chain rule states, dv/dx = dv/da * da/dx

dv/da = cosa, da/dx = -4

:. dv/dx = cosa * -4 = -4cosa

back to product rule which states:

dy/dx = V.du/dx + U.dv/dx

v = sin(-4x), u =3, du/dx = 0, dv/dx = -4cosa

therefore, dy/dx = sin(-4x) * 0 + 3 * -4cosa

-12cosa. Recall that a = -4x

dy/dx = -12cos(-4x)...


Abeg who knows a quicker method?

The 3 is a constant and not a function of x so you can't represent it with "u" neither should you use product rule. The traditional way of doing this is using the chain rule as shyneptune did. But as jamb candidates theres a shorter alternative.
We all know that if y = sinx, dy/dx = cos x
If y = tanx, dy/dx = sec²x and
If y = cos x, dy/dx = -sinx,
Fine, what if we have y = sin(ax),
To differentiate y, we differentiate ax (which is a) and multiply it by the differentiation of "sin" function which is "Cos". I.e
If y = sin(ax), dy/dx = cos(ax)

E.g., if y = Cos(3x² - 2), find dy/dx
Firstly, we find the differentiation of 3x² - 2 (which is 6x) and then we multiply it by the derivative of the "cos" function which is "-sin". Therefore,
dy/dx = 6x * (-sin)(3x² - 2)
= -6xsin(3x² - 2).

Then finally, back to the question in quote,
If y = b cos(ax), to find y´, we consider the b as just a constant (since we're differentiating with respect to x and b doesn't contain x) that will be multiplied by our answer after solving.
I.e., if y = b cos(ax),
dy/dx = b * -asin(ax) = -absin(ax)

So if y = 3 sin(-4x),
First and foremost, remember that 3 is just a constant, and that the differentiation of (-4x) is -4, then following the steps I earlier stated, the differentiation of sin(-4x) = -4cos(-4x) (remember that the differentiation of sinx function is Cosx).
Therefore, dy/dx = 3 * (-4cos(-4x))
= -12cos(-4x)

Another example,
Given y = 2tan5x,
dy/dx is simply = 2 * 5sec²5x
= 10sec²5x.

But if we have something like y = xsinx, this time around, we must use product rule since x is a function of x multiplied by another function of x, sinx. So u = x and v = sin x and so on....

If you can't explain, don't solve.

The probability of an event P is 3/4 while that of another event Q is 1/6. If the probability of both P and Q is 1/12, what is the probability of either P or Q?

Cc. Mathfaro, shyneptune, vickvan, orezy5, everyone.

Geofavor:
If you can't explain, don't solve.

The probability of an event P is 3/4 while that of another event Q is 1/6. If the probability of both P and Q is 1/12, what is the probability of either P or Q?

Cc. Mathfaro, shyneptune, vickvan, orezy5, everyone.

using P(PUQ)=P(P)+P(Q)-P(PnQ).
Continue from here

.

thankyouJesus:
using P(PUQ)=P(P)+P(Q)-P(PnQ). Continue from here

okay. The confusion i'm having is the set notation. It's been long

thankyouJesus:
using P(PUQ)=P(P)+P(Q)-P(PnQ). Continue from here

okay. The confusion i'm having is the set notation. It's been long

aj4emmanuel:
i'm nt dat gud in maths, but lemme just try dis one
(1/12)/(3/4) + (1/12)/(1/6)
(1/12*4/3) + (1/12*6/1)
1/9 + 1/2
11/18

try again

Geofavor:

okay. The confusion i'm having is the set notation. It's been long

P(PUQ) = probability of P or Q as requested for in the question
P(PnQ) = probability of both P and Q = 1/12
P(P)= probability of P = 3/4
P(Q) = probability of Q = 1/6
And since P and Q are not mutually exclusive, the probability of P or Q, P(PUQ)=P(P)+P(Q)-P(PnQ)
The ball is now in your court...

@ Umartins1 Geofavour Thankyoujesus Orezy5 mathefaro

vickvan:
@ Umartins1 Geofavour Thankyoujesus Orezy5 mathefaro

Algebraic substitution method will work.
Let u=2x+1.
Find du/dx = 2.
Dx=du/2.
Substitute back into the integral and evaluate.
Cc Mathefaro, I sight you bro.

Good morning guys

thankyouJesus:
Algebraic substitution method will work.
Let u=2x+1.
Find du/dx = 2.
Dx=du/2.
Substitute back into the integral and evaluate.
CcI sight you bro.

my Oga I remain loyal o

mathefaro:
my Oga I remain loyal o

Since we dey same hostel, we go jam for school na.
Have a wonderful day sir.

Another example here;
Note that the 1/3 in the first line of solution is the reciprocal of the differentiation of 3x - 5 (which is 3) and the 5 in the denominator is the same as the power got by increasing the initial power(4) by 1 (4 + 1)

thankyouJesus:
Since we dey same hostel, we go jam for school na.
Have a wonderful day sir.

you too bro

sub just expired,,,,, now that I need it most.

Y = Xsinx find dy/dx i solved this in about 40seconds. I feel it shouldn't take that long to solve.

is there a shortcut to solve questions like this rather than using the product rule?

Thankyoujesus, mathfaro

Geofavor:
Y = Xsinx find dy/dx i solved this in about 40seconds. I feel it shouldn't take that long to solve.

is there a shortcut to solve questions like this rather than using the product rule?

Thankyoujesus, mathfaro

yes there is but, care is needed (my phone is "very" bad, lets hope Mathefaro will point out that).
The short cut is,
y=xsinx.
To differentiate wrt x,
take x as a constant and differentiate sinx, introduce +, take sinx as constant and differentiate x.
Something like this,
x(cosx)+sinx(1)=xcosx+sinx.
My phone is bad, just try and understand.
Note: I and Mathefaro will be resuming very soon but, we will not desert here.

thankyouJesus:
yes there is but, care is needed (my phone is "very" bad, lets hope Mathefaro will point out that).
The short cut is,
y=xsinx.
To differentiate wrt x,
take x as a constant and differentiate sinx, introduce +, take sinx as constant and differentiate x.
Something like this,
x(cosx)+sinx(1)=xcosx+sinx.
My phone is bad, just try and understand.
Note: I and Mathefaro will be resuming very soon but, we will not desert here.

I understand. Hmm.

Let me form a question and solve.

If y = -3xcosx find dy/dx

dy/dx = -3x(-sinx) + -3(cosx)

= 3xsinx - 3cosx

lemme verify. Y = -3xcosx

v = -3x, u = cosx,

-3x(-sinx) + cosx(-3)

= 3xsinx -3cosx

alright.

I wish you and mathfaro excellence as you resume. But don't leave us yet o. Lol. This is when we need you most, as the exam is just a month away.

I'm gonna be needing tips on how to answer some questions fast (in seconds). Especially in physics and maths.

Thanks.

Geofavor:

I understand. Hmm.

Let me form a question and solve.

If y = -3xcosx find dy/dx

dy/dx = -3x(-sinx) + -3(cosx)

= 3xsinx - 3cosx

lemme verify. Y = -3xcosx

v = -3x, u = cosx,

-3x(-sinx) + cosx(-3)

= 3xsinx -cosx

alright.

I wish you and mathfaro excellence as you resume. But don't leave us yet o. Lol. This is when we need you most, as the exam is just a month away.

I'm gonna be needing tips on how to answer some questions fast (in seconds). Especially in physics and maths.

Thanks.

Thank you o, check the emboldened solution.

thankyouJesus:
Thank you o, check the emboldened solution.

... Number omission. Could be costly during exam.

Solve the inequality
1/2x + 1/4 > 1/3x + 1/2
mathfaro, thankyoujesus, shyneptune, etc

Geofavor:
Solve the inequality

1/2x + 1/4 > 1/3x + 1/2

mathfaro, thankyoujesus, shyneptune, etc

x<2/3 or x<0

.

Geofavor:

I understand. Hmm.

Let me form a question and solve.

If y = -3xcosx find dy/dx

dy/dx = -3x(-sinx) + -3(cosx)

= 3xsinx - 3cosx

lemme verify. Y = -3xcosx

v = -3x, u = cosx,

-3x(-sinx) + cosx(-3)

= 3xsinx -3cosx

alright.

I wish you and mathfaro excellence as you resume. But don't leave us yet o. Lol. This is when we need you most, as the exam is just a month away.

I'm gonna be needing tips on how to answer some questions fast (in seconds). Especially in physics and maths.

Thanks.

I almost didn't get the mention. You've omitted the "e" on my handle

Geofavor:
Solve the inequality

I think someone has solved something already. are you a jambite or an undergraduate?

aj4emmanuel:
(2+x)/4x > (2+3x)/6x
6x(2+x) > 4x(2+3x)
12x + 6x^2 > 8x + 12x^2
0 > 12x^2 - 6x^2 + 8x - 12x
0 > 6x^2 - 4x
6x^2 - 4x <0
(2x)(3x-2) <0
2x <0 or 3x - 2 <0
x <0 ,3x <2
x <0 , x <2/3
2/3 >x< 0

Greatly done, nice work but, I think the last part should be 0>x<2/3 (reason because 2/3 is greater than 0)

Actually, I think you guys have been looking at the question from a wrong angle. Something tells me that he meant (1/2)x or x/2 and (1/3)x or x/3 instead of the 1/2x and 1/3x respectively in the question.
My thoughts though