Question help!

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!

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)

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?

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.

$\dfrac{1}{t^{1/3}} = t^{-1/3}$
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...

Great

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

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