Yatayyat
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#1
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!
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BobbJo
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#2
(Original post by 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!
multiply top/bottom by 3t^(1/3)
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RDKGames
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(Original post by 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!
But there is no x 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 t^{5/3}.

Then compare your form to theirs. They have a 3t^{1/3} in the denominator, so multiply your form's numerator and denominator by it... then simplify the numerator.
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Yatayyat
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#4
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(Original post by BobbJo)
multiply top/bottom by 3t^(1/3)
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?
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BobbJo
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(Original post by 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?
\dfrac{1}{t^{1/3}} = t^{-1/3}
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
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Yatayyat
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(Original post by RDKGames)
But there is no x 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 t^{5/3}.

Then compare your form to theirs. They have a 3t^{1/3} in the denominator, so multiply your form's numerator and denominator by it... then simplify the numerator.
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) (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
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Yatayyat
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(Original post by BobbJo)
\dfrac{1}{t^{1/3}} = t^{-1/3}
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
I just spotted that I soon as soon as you replied. Thank you! I think I might have got it now...
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BobbJo
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(Original post by Yatayyat)
I just spotted that I soon as soon as you replied. Thank you! I think I might have got it now...
Great

Here is a link if you want to learn to use latex: https://www.thestudentroom.co.uk/help/latex
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Yatayyat
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(Original post by BobbJo)
Great

Here is a link if you want to learn to use latex: https://www.thestudentroom.co.uk/help/latex
Thanks Will check it out
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RDKGames
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(Original post by 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.
Looks good, denominator should say cube root(t) tho
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Yatayyat
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(Original post by RDKGames)
Looks good, denominator should say cube root(t) tho
yeah that was a accident typo, just changed it now. Thank you!
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