HSC 2016 Maths Marathon (archive) (3 Viewers)

Nailgun · Jan 13, 2016

Re: HSC 2016 2U Marathon

erm I never really understood that notation, that's why I kinda just use the y'
It worked so far so I guess I just went with it
Why would you need to differentiate the (sinx)^n though? Wouldnt you end up with $(\sin { x } )^{ n }\pi ^{ (\sin { x } )^{ n }-1 }$

btw im not trying to argue or anything im just a bit confused

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

$Let\quad y={ \pi }^{ \sin ^{ n }{ x } }\\ Then\quad let\quad u=\sin ^{ n }{ x } \Rightarrow \frac { du }{ dx } =n\cos { x } \sin ^{ n-1 }{ x } \\ y={ \pi }^{ u }\Rightarrow \frac { dy }{ du } ={ \pi }^{ u }\ln { \pi } \\ \frac { dy }{ dx } =\frac { dy }{ du } \frac { du }{ dx } \\ \therefore \frac { d }{ dx } \left( { \pi }^{ \sin ^{ n }{ x } } \right) =\left( { \pi }^{ u } \ln { \pi } \right) \left( n\cos { x } \sin ^{ n-1 }{ x } \right) \\ \frac { d }{ dx } \left( { \pi }^{ \sin ^{ n }{ x } } \right) =n\ln { \pi } \, { \pi }^{ sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x }$

I think you completely missed the fact that x is the variable we're differentiating with respect to. Not the entire sin(x) quantity.

So the derivative of the inside itself is already tedious enough, and then you have to multiply the little extra of the derivative of the whole thing.

http://www.wolframalpha.com/input/?i=d/dx+pi^((sinx)^n)

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

$\\ \frac { d }{ dx } \left( { e }^{ \sin ^{ n }{ x } } \right) =n{ e }^{ \sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x } \\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }=n\left( n{ \pi }^{ \sin ^{ n }{ x } }\ln { \pi } \cos { x } \sin ^{ n-1 }{ x } +n{ e }^{ \sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }$

$\\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }={ n }^{ 2 }\left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }\\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }\cdot \cos { \left( nx \right) } ={ n }^{ 2 }\cos { \left( nx \right) } \left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }$

Because the dot technically ends the functionality of the derivative operator. Need a bracket around the entire expression for it to become a product rule question

InteGrand · Jan 13, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\\ \frac { d }{ dx } \left( { e }^{ \sin ^{ n }{ x } } \right) =n{ e }^{ \sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x } \\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }=n\left( n{ \pi }^{ \sin ^{ n }{ x } }\ln { \pi } \cos { x } \sin ^{ n-1 }{ x } +n{ e }^{ \sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }$

$\\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }={ n }^{ 2 }\left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }\\ \frac { d }{ dx } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }\cdot \cos { \left( nx \right) } ={ n }^{ 2 }\cos { \left( nx \right) } \left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }$

Because the dot technically ends the functionality of the derivative operator. Need a bracket around the entire expression for it to become a product rule question

Guessing he (Drsoccerball) meant it to be a product rule and typo'ed, since there's not much point just randomly putting a factor next to it for no reason, as it'd just be "carried along for the ride", so to speak.

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Guessing he (Drsoccerball) meant it to be a product rule and typo'ed, since there's not much point just randomly putting a factor next to it for no reason, as it'd just be "carried along for the ride", so to speak.

$\frac { d }{ dx } \left( { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }\cos { \left( nx \right) } \right) ={ n }^{ 2 }\cos { \left( nx \right) } \left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }-n\sin { \left( nx \right) } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }$

Too long. Lazy to simplify and factor out the ...^(n-1)

InteGrand · Jan 13, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\frac { d }{ dx } \left( { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }\cos { \left( nx \right) } \right) ={ n }^{ 2 }\cos { \left( nx \right) } \left( { \pi }^{ \sin ^{ n }{ x } }\ln { \pi } +{ e }^{ \sin ^{ n }{ x } } \right) \left( \cos { x } \sin ^{ n-1 }{ x } \right) { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n-1 }-n\sin { \left( nx \right) } { \left( { \pi }^{ \sin ^{ n }{ x } }+{ e }^{ \sin ^{ n }{ x } } \right) }^{ n }$

Too long. Lazy to simplify and factor out the ...^(n-1)

And his original question was to find the second derivative.

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
And his original question was to find the second derivative.

If I was bothered

Drsoccerball · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
erm I never really understood that notation, that's why I kinda just use the y'
It worked so far so I guess I just went with it
Why would you need to differentiate the (sinx)^n though? Wouldnt you end up with $(\sin { x } )^{ n }\pi ^{ (\sin { x } )^{ n }-1 }$

btw im not trying to argue or anything im just a bit confused

The way you differentiate exponentials are different to any other function you are differentiating. For this type of question instead of wasting a few lines with a u substitution you can do this:

$y= \pi ^{\sin ^nx}$

$\ln{y}= \ln{\pi} \cdot {\sin ^ nx}$ ln both sides and use the 'burning lady log rule' or the 'broken pipette rule' (yeah don't ask...) $$

$\frac{y'}{y}= \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x $ Differentiate both sides. $$

$y'= y \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

$\therefore y' = \pi ^{\sin ^nx}\ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
The way you differentiate exponentials are different to any other function you are differentiating. For this type of question instead of wasting a few lines with a u substitution you can do this:

$y= \pi ^{\sin ^nx}$

$\ln{y}= \ln{\pi} \cdot {\sin ^ nx}$ ln both sides and use the 'burning lady log rule' or the 'broken pipette rule' (yeah don't ask...) $$

$\frac{y'}{y}= \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x $ Differentiate both sides. $$

$y'= y \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

$\therefore y' = \pi ^{\sin ^nx}\ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

$\noindent For any curious students, this technique is known as \textit{logarithmic differentiation}. It's especially useful if the base is a function of $x$. Here's an example of a question where logarithmic differentiation can be helpful.$

$\noindent \textbf{ANOTHER QUESTION}$

$\noindent Use logarithmic differentiation to find $\frac{\mathrm{d}y}{\mathrm{d}x}$, where $y = x^{\sin x}$, defined for $x>0$.$

leehuan · Jan 14, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
The way you differentiate exponentials are different to any other function you are differentiating. For this type of question instead of wasting a few lines with a u substitution you can do this:

$y= \pi ^{\sin ^nx}$

$\ln{y}= \ln{\pi} \cdot {\sin ^ nx}$ ln both sides and use the 'burning lady log rule' or the 'broken pipette rule' (yeah don't ask...) $$

$\frac{y'}{y}= \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x $ Differentiate both sides. $$

$y'= y \ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

$\therefore y' = \pi ^{\sin ^nx}\ln{\pi} \cdot n\cos{x} \sin ^{n-1}x$

Note that had InteGrand not explained it, this is something a 2U student would never have thought about as it is far beyond the scope of the course. It's pretty much manipulating implicit differentiation.

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$Let\quad y={ \pi }^{ \sin ^{ n }{ x } }\\ Then\quad let\quad u=\sin ^{ n }{ x } \Rightarrow \frac { du }{ dx } =n\cos { x } \sin ^{ n-1 }{ x } \\ y={ \pi }^{ u }\Rightarrow \frac { dy }{ du } ={ \pi }^{ u }\ln { \pi } \\ \frac { dy }{ dx } =\frac { dy }{ du } \frac { du }{ dx } \\ \therefore \frac { d }{ dx } \left( { \pi }^{ \sin ^{ n }{ x } } \right) =\left( { \pi }^{ u } \ln { \pi } \right) \left( n\cos { x } \sin ^{ n-1 }{ x } \right) \\ \frac { d }{ dx } \left( { \pi }^{ \sin ^{ n }{ x } } \right) =n\ln { \pi } \, { \pi }^{ sin ^{ n }{ x } }\cos { x } \sin ^{ n-1 }{ x }$

I think you completely missed the fact that x is the variable we're differentiating with respect to. Not the entire sin(x) quantity.

So the derivative of the inside itself is already tedious enough, and then you have to multiply the little extra of the derivative of the whole thing.

http://www.wolframalpha.com/input/?i=d/dx+pi^((sinx)^n)

Okay that actually helped heaps, I've never really thought of the chain rule as a formal process, it was kind of taught to me as multiply by the power and the derivative inside the brackets and subtract 1 from the power
Basically the top part or dy is what the original equation is equal to? And the bottom is what you are differentiating with respect too right
Chain rule is then splitting up an expression into a u part and a y part and then you can differentiate the whole yeah
Imma go back and try redo the question

Also the logarithic differentiation question (integrand)
$y={ x }^{ \sin { x } }\\ \ln { y } =\ln { { x }^{ \sin { x } } } \\ \ln { y } =\ln { x } \times \sin { x } \\ \frac { y' }{ y } =\frac { \sin { x } }{ x } +\ln { x } \cos { x } \\ y'=y(\frac { \sin { x } }{ x } +\ln { x } \cos { x } )\\ y'={ x }^{ \sin { x } }(\frac { \sin { x } }{ x } +\ln { x } \cos { x } )\\ y'={ x }^{ \sin { x } -1 }\sin { x } +\ln { x } \cos { x } { x }^{ \sin { x } }$

EDIT: Having a mindblown moment now that the Leibniz notation actually makes some sense

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Also the logarithic differentiation question (integrand)
$y={ x }^{ \sin { x } }\\ \ln { y } =\ln { { x }^{ \sin { x } } } \\ \ln { y } =\ln { x } \times \sin { x } \\ \frac { y' }{ y } =\frac { \sin { x } }{ x } +\ln { x } \cos { x } \\ y'=y(\frac { \sin { x } }{ x } +\ln { x } \cos { x } )\\ y'={ x }^{ \sin { x } }(\frac { \sin { x } }{ x } +\ln { x } \cos { x } )\\ y'={ x }^{ \sin { x } -1 }\sin { x } +\ln { x } \cos { x } { x }^{ \sin { x } }$

$\noindent Correct!$

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

So is In y = In x always or are there like conditions?
Thats very useful though

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
So is In y = In x always or are there like conditions?
Thats very useful though

$\noindent Not sure what you mean exactly. We don't say $\ln y = \ln x$ (since that'd be true if and only if $y=x$). Rather, we do $\ln y = \ln \left( f(x)\right)$, where $y=f(x)$.$

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

yeah thats what I meant
In y = In f(x)
Where y = f(x)

Which just made me realise that they're the same thing already
What was the special rule then? y'/y?

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
yeah thats what I meant
In y = In f(x)
Where y = f(x)

Which just made me realise that they're the same thing already
What was the special rule then? y'/y?

$\noindent Yeah, it's basically this rule: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y \right) = \frac{y^\prime}{y}=\frac{f^\prime (x)}{f(x)}$, where $y=f(x)$. This comes from the chain rule.$

Nailgun · Jan 14, 2016

Re: HSC 2016 2U Marathon

Oh right, I was aware of that since it was in my textbook, although it was never really used like that
I know its basically just another application of it, but generally like
$y=\ln { ({ x }^{ 2 }+1) } \\ \frac { dy }{ dx } =\frac { 2x }{ { x }^{ 2 }+1 }$

I wouldn't of thought to take the log of both sides which is why I thought it was the special part lol

leehuan · Jan 14, 2016

Re: HSC 2016 2U Marathon

Implicit differentiation is another direct consequence of the chain rule:

$\\ \frac { d }{ dx } { \left( f\left( x \right) \right) }^{ n }=nf^{ \prime }\left( x \right) { \left( f\left( x \right) \right) }^{ n-1 }\\ \frac { d }{ dx } { y }^{ n }=n{ y }^{ n-1 }\frac { dy }{ dx } \\ \\ \frac { d }{ dx } \ln { f\left( x \right) } =\frac { f^{ \prime }\left( x \right) }{ f\left( x \right) } \\ \frac { d }{ dx } \ln { y } =\frac { \frac { dy }{ dx } }{ y } \\ \\$

They call that logarithmic differentiation as a special case of it simply because of the ability to manipulate the power rule of logarithms.

Of course, and also because the process introduces a log temporarily itself.

The rule is somewhat broken in the event you have something like y=-x^2 and it's all negative.

InteGrand · Jan 14, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Of course, and also because the process introduces a log temporarily itself.

The rule is somewhat broken in the event you have something like y=-x^2 and it's all negative.

$\noindent Actually, it still works if we blindly apply our familiar log and algebra laws.$

leehuan · Jan 14, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$\noindent Actually, it still works if we blindly apply our familiar log and algebra laws.$

Yep. I can see that. But some may still just yell at you for introducing a negative lol

Troll way around -> d/dx ln|x| = 1/x

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