Please Help urgently!!!!! MATHS

[rehan751]

WEll in this exercise we differentiate those equations that have a whole power of a larger number.

Q.1 (c) is this: (¼x + 2)^5

To differentiate this, follow these steps:

1. write down the power of the bracket as a coefficient in your answer: e.g write 5 for this question.

2. Then write the bracket AS IT IS, but subtract 1 from the power and write the bracket down with its subtracted power: e.g (¼x + 2)^4

We had '5' from the first step. So it will make it like this : 5 * (¼x + 2)^4

3. Finally differentiate the inside of the bracket like you used to do in 15.1 and write it next: e.g: ¼

So the answer will be like this: 5 * (¼x + 2)^4 * ¼

= 5/4 * (¼x + 2)^4

Hope you might understand it

P.S * stands for multiply
¼ is 1/4

 

[SalmanPakRocks]

Rehan is right !!

 

[scouserlfc]

Oh i get it this is the same thing as 15.1 but you have to add another bracket without differentiating its inside I get it even if what I wrote looks as if it does not make sense. :lol:

 

[SalmanPakRocks]

But its right right ??

 

[rehan751]

So scouserlfc did you understand it?

 

[scouserlfc]

YES I DID AND SORRY I AGAIN WANT YOUR HELP FOR 15.3 OF THE SAME EXERCISE HOW DO WE DO THIS EXPLAIN IT BY DOIN QUESTION 1 PART f OF THE EXERCISE 15.3.THANKS.,

 

[rehan751]

In this exercise there are two functions of x ( In simpler term a function of x is that which includes x in any power)

For example in this question Q.1 (f) it is: x (1-x^2)^2

So the first function is : x
2nd function is : (1-x^2)^2

Lets call the first function 'u'
Lets call the 2nd function 'v'

To solve these type of questions there is a formula: dy/dx = u v' + v u' ( v' means derivative of v or dy/dx of v, similarly u' means derivative of u)

Derivative of u is: 1
Derivative of v is: 2 * (1-x^2) * -x
: -2x * (1-x^2)
: -2x + x^3

Just put these in the formula: u v' + v u'

dy/dx = x ( -2x + x^3) + (1-x^2)^2

= -2x^2 + x^4 + 1 - 2x^2+ x^4 [ I opened (1-x^2)^2 ]

= -4x^2 + 2x^4 + 1

[[[[ Well the answer is different at the back of the book ]]]]

Just get the method maybe I've made some mistakes in the working ( Its been a long time Ive done this exercise so I am bound to make mistakes )

 

[rehan751]

Let me do the first part of question 1 to explain ( since I got the above answer wrong)

Q.1 (a) is: (x-1) (x+2)^2

Derivative of u: 1

Derivative of v: 2 (x+2)

dy/dx = u * v' + v * u'

= (x-1) (2) (x+2) + (x+2)^2 (1)

= 2 ( x^2 + 2x -x -2) + x^2 + 4x + 4

= 2x^2 + 2x -4 + x^2 + 4x + 4

= 3x^2 + 6x
Ans

 

[scouserlfc]

thanks man well i think i spotted your mistake try to find your self well let me see if this works.

 

[SalmanPakRocks]

 

[scouserlfc]

WELL YEAH THANKS TO @ REHAN751 THAT HE MADE ME UNDERSTAND THIS THANKS MATE.

AND YEAH THANKS TO @abcde and @salmanpakrocks for HELPING ME OUT IN THE CIRCULAR GEOMETRY I GOT AN A* IN MY SCHOOL EXAM THANKS TO YOU GUYS .JAZAKALLAH TO ALL OF YOU WHO HAD HAVE HELPED ME...

 

[SalmanPakRocks]

wow great!! Well no problem and yeah best of luck in the CIE !!

 

[scouserlfc]

Yeah thanks and good luck to you guys as well.

 

[SalmanPakRocks]

thanks !!

 

[CaptainDanger]

Buy Maths Key Points! Everything is explained in it!

 

[scouserlfc]

Yeah thanks but ill get this in the next year as i am only giving P.Studies,Urdu and Islamiat this year and i just wanna get my school work done as quick as possible.Yup thanks for the suggestion.

 

[SalmanPakRocks]

Here you go scouserlfc I think this will help you in maths

 

[scouserlfc]

hEY DID U MAKE THESE NOTES OR YOU GOT IT FROM THE COMPETITION SECTION.WHOEVER MADE THEM DESERVES A BIG THUMBS UP.BY THE WAY THANKS FOR HELPN AGAIN.

 

[SalmanPakRocks]

No i didnot made these! I got it from the competition section but as you needed help i provided them to you !

 

[scouserlfc]

THANKS >