2U HSC Questions Thread (1 Viewer)

gummibear0731 · Oct 15, 2016

I have to differentiate: 3x/√3x-4+√3x-4

I end up getting (9√2)/8 but the answer is (3√2)8

I'm so close yet so far... help!

InteGrand · Oct 15, 2016

gummibear0731 said:
I have to differentiate: 3x/√3x-4+√3x-4

I end up getting (9√2)/8 but the answer is (3√2)8

I'm so close yet so far... help!

$\noindent What is the function (need brackets, it's hard to tell what it is)? Also, is it saying to evaluate the derivative at a particular point? If so, where?$

gummibear0731 · Oct 15, 2016

Sorry, I wasn't very clear. I have attached a screenshot of the question!

InteGrand · Oct 15, 2016

gummibear0731 said:
Sorry, I wasn't very clear. I have attached a screenshot of the question!View attachment 33504 View attachment 33504

$\noindent We have $f(x) = x\sqrt{3x-4} = x\left(3x-4\right)^{\frac{1}{2}}$. Recall that $\left(uv\right)'' = u''v + 2u'v' + uv''$ for twice-differentiable functions $u,v$. Note that $x'' = 0, x' = 1, v' = \frac{3}{2}\left(3x-4\right)^{-\frac{1}{2}}$ and $v'' = - \frac{9}{4}\left(3x-4\right)^{-\frac{3}{2}}$, where $v = \left(3x-4\right)^{\frac{1}{2}}$. Hence$

$\begin{align*}f'' (x) &= 2v' + xv'' \\ &= 3\left(3x-4\right)^{-\frac{1}{2}} - \frac{9}{4}x\left(3x-4\right)^{-\frac{3}{2}}.\end{align*}$

$\noindent Note that $3\times 2 - 4 = 2$. Hence$

$\begin{align*}f''(2) &= 3 \times 2 ^{-\frac{1}{2}} - \frac{9}{4}\times 2\times 2^{-\frac{3}{2}}\\ &= \frac{3\sqrt{2}}{2} - \frac{9}{4}\times 2\times \frac{1}{2}\times \frac{\sqrt{2}}{2}\\ &= \left(\frac{3}{2} - \frac{9}{8}\right)\sqrt{2}\\ &= \frac{3\sqrt{2}}{8}. \end{align*}$

$\noindent \textbf{Note.} To see why $(uv)'' = u''v + 2u'v' + uv''$ for twice-differentiable functions $u,v$, just apply the product rule twice:$

$\begin{align*}(uv)'' &= \left((uv)'\right)' \\ &= (u'v + uv')'\\ &= \left(u'v\right)' + \left(uv'\right)' \\ &= u''v + u'v' + u'v' + uv'' \quad (\text{product rule}) \\ &= u''v + 2u'v' + uv''. \end{align*}$

$\noindent In fact, in general for $n$-times differentiable functions $u,v$, the $n$-th derivative of $uv$ is $(uv)^{\left(n\right)} = \sum_{k = 0}^{n}\binom{n}{k}u^{(n-k)}v^{(k)}$, which one can show by induction.$

gummibear0731 · Oct 15, 2016

I'm sure you're right... I haven't learnt that rule yet though... do we need to know it for year 12?

InteGrand · Oct 15, 2016

gummibear0731 said:
I'm sure you're right... I haven't learnt that rule yet though... do we need to know it for year 12?

No, you don't need to know it for Year 12. You could (and are expected to) just find the second derivative of the function normally (i.e. find the first derivative first, then differentiate again). It's just a bit more tedious to do it that way I think.

2U HSC Questions Thread (1 Viewer)

gummibear0731

New Member

InteGrand

Well-Known Member

gummibear0731

New Member

InteGrand

Well-Known Member

gummibear0731

New Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)