# Question help!

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

https://imgur.com/rY9WJEq

It's part a that I'm stuck on!

What I have done so far is that I have realised that I must use the quotient rule to differentiate the expression because the top and bottom of the fraction are both functions of x.

So to diff w.r.t x, I have let u = 4t^2/3 and v = t^2 + 1, then I get du/dt = 8/3 t^-1/3 and dv/dt = 2t.

Then substituting all that into the quotient formula I get:

[(t^2 + 1) * (8/3 t^-1/3) - (4t^2/3) * (2t)] / (t^2+1)^2

Then multiplying out the brackets at the top of the fraction, it gives me,

[(8/3 t^5/3) + (8/3t^-1/3) - (8t^5/3)] / (t^2+1)^2.

I'm not sure what the steps would be to break down the differentiated expression into the required form given in the question?

Any help would be appreciated. Thanks!

It's part a that I'm stuck on!

What I have done so far is that I have realised that I must use the quotient rule to differentiate the expression because the top and bottom of the fraction are both functions of x.

So to diff w.r.t x, I have let u = 4t^2/3 and v = t^2 + 1, then I get du/dt = 8/3 t^-1/3 and dv/dt = 2t.

Then substituting all that into the quotient formula I get:

[(t^2 + 1) * (8/3 t^-1/3) - (4t^2/3) * (2t)] / (t^2+1)^2

Then multiplying out the brackets at the top of the fraction, it gives me,

[(8/3 t^5/3) + (8/3t^-1/3) - (8t^5/3)] / (t^2+1)^2.

I'm not sure what the steps would be to break down the differentiated expression into the required form given in the question?

Any help would be appreciated. Thanks!

0

reply

Report

#2

(Original post by

https://imgur.com/rY9WJEq

It's part a that I'm stuck on!

What I have done so far is that I have realised that I must use the quotient rule to differentiate the expression because the top and bottom of the fraction are both functions of x.

So to diff w.r.t x, I have let u = 4t^2/3 and v = t^2 + 1, then I get du/dt = 8/3 t^-1/3 and dv/dt = 2t.

Then substituting all that into the quotient formula I get:

[(t^2 + 1) * (8/3 t^-1/3) - (4t^2/3) * (2t)] / (t^2+1)^2

Then multiplying out the brackets at the top of the fraction, it gives me,

[(8/3 t^5/3) + (8/3t^-1/3) - (8t^5/3)] / (t^2+1)^2.

I'm not sure what the steps would be to break down the differentiated expression into the required form given in the question?

Any help would be appreciated. Thanks!

**Yatayyat**)https://imgur.com/rY9WJEq

It's part a that I'm stuck on!

What I have done so far is that I have realised that I must use the quotient rule to differentiate the expression because the top and bottom of the fraction are both functions of x.

So to diff w.r.t x, I have let u = 4t^2/3 and v = t^2 + 1, then I get du/dt = 8/3 t^-1/3 and dv/dt = 2t.

Then substituting all that into the quotient formula I get:

[(t^2 + 1) * (8/3 t^-1/3) - (4t^2/3) * (2t)] / (t^2+1)^2

Then multiplying out the brackets at the top of the fraction, it gives me,

[(8/3 t^5/3) + (8/3t^-1/3) - (8t^5/3)] / (t^2+1)^2.

I'm not sure what the steps would be to break down the differentiated expression into the required form given in the question?

Any help would be appreciated. Thanks!

0

reply

Report

#3

**Yatayyat**)

https://imgur.com/rY9WJEq

It's part a that I'm stuck on!

What I have done so far is that I have realised that I must use the quotient rule to differentiate the expression because the top and bottom of the fraction are both functions of x.

So to diff w.r.t x, I have let u = 4t^2/3 and v = t^2 + 1, then I get du/dt = 8/3 t^-1/3 and dv/dt = 2t.

Then substituting all that into the quotient formula I get:

[(t^2 + 1) * (8/3 t^-1/3) - (4t^2/3) * (2t)] / (t^2+1)^2

Then multiplying out the brackets at the top of the fraction, it gives me,

[(8/3 t^5/3) + (8/3t^-1/3) - (8t^5/3)] / (t^2+1)^2.

I'm not sure what the steps would be to break down the differentiated expression into the required form given in the question?

Any help would be appreciated. Thanks!

Anyway, I think by wrt x you meant wrt t. Also if you can use LaTeX to write your equations they would be much easier to read!

You can simplify the numerator since there are two terms with .

Then compare your form to theirs. They have a in the denominator, so multiply your form's numerator and denominator by it... then simplify the numerator.

0

reply

(Original post by

multiply top/bottom by 3t^(1/3)

**BobbJo**)multiply top/bottom by 3t^(1/3)

But how does the times t^1/3 help?

1

reply

Report

#5

(Original post by

I understand now that multiplying by 3 gets rid of the fractions at the top, so I now get [8t^5/3 + 8t^-1/3 - 24t^5/3] / 3(t^2+1)^2.

But how does the times t^1/3 help?

**Yatayyat**)I understand now that multiplying by 3 gets rid of the fractions at the top, so I now get [8t^5/3 + 8t^-1/3 - 24t^5/3] / 3(t^2+1)^2.

But how does the times t^1/3 help?

so the term above is also a fraction

we get rid of the fractional powers of t in the numerator, leading to the form in the given answer

Last edited by BobbJo; 1 year ago

0

reply

(Original post by

But there is no here...

Anyway, I think by wrt x you meant wrt t. Also if you can use LaTeX to write your equations they would be much easier to read!

You can simplify the numerator since there are two terms with .

Then compare your form to theirs. They have a in the denominator, so multiply your form's numerator and denominator by it... then simplify the numerator.

**RDKGames**)But there is no here...

Anyway, I think by wrt x you meant wrt t. Also if you can use LaTeX to write your equations they would be much easier to read!

You can simplify the numerator since there are two terms with .

Then compare your form to theirs. They have a in the denominator, so multiply your form's numerator and denominator by it... then simplify the numerator.

So I'm guessing after multiplying top and bottom by 3, to simplify the top, you can factor out t^-1/3 as this is common to all terms at the top.

I get 8t^-1/3 (t^2 + 1 -3t^2)/ 3(t^2+1)^2

Then it's apparent that 8t^-1/3 (1 - 2t^2) / 3(t^2+1)^2.

The 't^-1/3' at the numerator can come down, so it can then be 8(1-2t^2)/ 3* cube root(t) (t^2+1)^2.

Really sorry for the poor formatting, I don't know how to use LaTeX, so should probably learn how to use it soon.

Last edited by Yatayyat; 1 year ago

0

reply

(Original post by

so the term above is also a fraction

we get rid of the fractional powers of t in the numerator, leading to the form in the given answer

**BobbJo**)so the term above is also a fraction

we get rid of the fractional powers of t in the numerator, leading to the form in the given answer

0

reply

Report

#8

(Original post by

I just spotted that I soon as soon as you replied. Thank you! I think I might have got it now...

**Yatayyat**)I just spotted that I soon as soon as you replied. Thank you! I think I might have got it now...

Here is a link if you want to learn to use latex: https://www.thestudentroom.co.uk/help/latex

0

reply

(Original post by

Great

Here is a link if you want to learn to use latex: https://www.thestudentroom.co.uk/help/latex

**BobbJo**)Great

Here is a link if you want to learn to use latex: https://www.thestudentroom.co.uk/help/latex

0

reply

Report

#10

(Original post by

Yes you're right, it was meant to be 'wrt t', sorry.

So I'm guessing after multiplying top and bottom by 3, to simplify the top, you can factor out t^-1/3 as this is common to all terms at the top.

I get 8t^-1/3 (t^2 + 1 -3t^2)/ 3(t^2+1)^2

Then it's apparent that 8t^-1/3 (1 - 2t^2) / 3(t^2+1)^2.

The 't^-1/3' at the numerator can come down, so it can then be 8(1-2t^2)/ 3* cube root(t^2+1)^2.

Really sorry for the poor formatting, I don't know how to use LaTeX, so should probably learn how to use it soon.

**Yatayyat**)Yes you're right, it was meant to be 'wrt t', sorry.

So I'm guessing after multiplying top and bottom by 3, to simplify the top, you can factor out t^-1/3 as this is common to all terms at the top.

I get 8t^-1/3 (t^2 + 1 -3t^2)/ 3(t^2+1)^2

Then it's apparent that 8t^-1/3 (1 - 2t^2) / 3(t^2+1)^2.

The 't^-1/3' at the numerator can come down, so it can then be 8(1-2t^2)/ 3* cube root(t^2+1)^2.

Really sorry for the poor formatting, I don't know how to use LaTeX, so should probably learn how to use it soon.

0

reply

(Original post by

Looks good, denominator should say cube root(t) tho

**RDKGames**)Looks good, denominator should say cube root(t) tho

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top