HSC 2016 MX1 Marathon - Harder 3U Integration (archive) (1 Viewer)

KINGOM 885 · Mar 13, 2016

This thread is for interesting integrals or derivatives or integrals and derivatives that you need help with. The only rules are that you must keep this to harder 3U and every question must be answered before a new one is asked. Here is one to start you off,
differentiate y= 7^ln(x) with respect to x

Nailgun · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

KINGOM 885 said:
This thread is for interesting integrals or derivatives or integrals and derivatives that you need help with. The only rules are that you must keep this to harder 3U and every question must be answered before a new one is asked. Here is one to start you off,
differentiate y= 7^ln(x) with respect to x

erm no offence but this is a 2U question and not a very hard one at that lele

$y=7\ln { x } \\ \frac { dy }{ dx } =\frac { 7 }{ x }$

jathu123 · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

Nailgun said:
erm no offence but this is a 2U question and not a very hard one at that lele

$y=7\ln { x } \\ \frac { dy }{ dx } =\frac { 7 }{ x }$

haha I think you forgot to read the exponential sign lol, he meant, $y=7^{ln(x)}$

Nailgun · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

jathu123 said:
haha I think you forgot to read the exponential sign lol, he meant, $y=7^{ln(x)}$

rekt

jathu123 · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

$y=7^{ln(x)} \\ ln(y)=ln(x)ln(7) \\ \frac{1}{y}\frac{dy}{dx} = ln(7)\frac{1}{x}\\ \frac{dy}{dx}=\frac{yln(7)}{x} \\ \frac{dy}{dx}= \frac{ln(7)7^{ln(x)}}{x}$

I'm not sure if this is the correct way of doing it though

Nailgun · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

$y=7^{ \ln { x } }\\ \ln { y } =\ln { 7^{ \ln { x } } } \\ \ln { y } =\ln { x } .\ln { 7 } \\ \frac { d }{ dx } \ln { y } =\frac { d }{ dx } \ln { x } .\ln { 7 } \\ \frac { y' }{ y } =\frac { \ln { 7 } }{ x } \\ y'=7^{ \ln { x } }.\frac { \ln { 7 } }{ x }$

edit: oops ninjaed

Nailgun · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

jathu123 said:
$y=7^{ln(x)} \\ ln(y)=ln(x)ln(7) \\ \frac{1}{y}\frac{dy}{dx} = ln(7)\frac{1}{x}\\ \frac{dy}{dx}=\frac{yln(7)}{x} \\ \frac{dy}{dx}= \frac{ln(7)7^{ln(x)}}{x}$

I'm not sure if this is the correct way of doing it though

You can also use chain rule, but yeah

leehuan · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

This seems boring...

$\\ \frac{d}{dx} \frac{{\left({e}^{x}\sin{x}\right)}^{\cos^{-1}{x}\sin^{-1}{x}}}{\ln{\left({x}^{2}+3\right)}}$

leehuan · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

Chain rule method is easy to use for the above derivative however I'm going to assume a result that you can't assume in the HSC

d/dx 7^ln(x)
= ln(7).7^ln(x).1/x

Assumed: d/dx a^x = a^x.ln(a)

Nailgun · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

leehuan said:
Chain rule method is easy to use for the above derivative however I'm going to assume a result that you can't assume in the HSC

d/dx 7^ln(x)
= ln(7).7^ln(x).1/x

Assumed: d/dx a^x = a^x.ln(a)

wait you can't assume this lol?

leehuan · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

Nailgun said:
wait you can't assume this lol?

Fortunately I am yet to seen an exam paper that actually asks you to take the derivative of a^x.

However, as the HSC gets further and further dumbed down, at least since the year 2010 (and of course actual HSC papers to 2001) I have not seen any strict mention to the fact that d/dx a^x = a^x.ln(x) is an assumed result. The syllabus is so outdated I don't bother with it.

Proof of the result:
d/dx a^x
= d/dx e^(ln(a^x)) - true as a^x>0 for all x in R
= d/dx e^(x.ln(a))
= ln(a).e^(x.ln(a))
reverse the steps
= ln(a).a^x

KINGOM 885 · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

That is correct. I personally made x the subject and used dx/dy to differentiate before reverting to dy/dx.

Paradoxica · Mar 13, 2016

Re: MX1 Calculus Marathon 2016

$\int \frac{\cos{x}-\sin{x}}{2+\sin{2x}}\,$d$x$

RivalryofTroll · Mar 14, 2016

Re: MX1 Calculus Marathon 2016

leehuan · Mar 14, 2016

Re: MX1 Calculus Marathon 2016

Hint for this one since no substitution is here.

Paradoxica said:
$\int \frac{\cos{x}-\sin{x}}{2+\sin{2x}}\,$d$x$

What does 1 = ?

jathu123 · May 19, 2016

Re: MX1 Calculus Marathon 2016

Paradoxica said:
$\int \frac{\cos{x}-\sin{x}}{2+\sin{2x}}\,$d$x$

$\noindent \int \frac{\cos x -\sin x}{2+\sin 2x} $ d$x \\ \\ =\int \frac{\cos x -\sin x}{1+(1+\sin 2x)} $ d$x\\ \\ = \int \frac{\cos x -\sin x}{1+(\sin ^{2}x+2\sin x\cos x+\cos^{2}x) } $ d$x \\ \\ = \int \frac{\cos x -\sin x}{1+(\sin x+\cos x)^{2}}$ d$x \\ \\ =\tan^{-1}(\sin x+\cos x ) + C$

si2136 · May 19, 2016

Re: MX1 Calculus Marathon 2016

Nailgun said:
wait you can't assume this lol?

You can?

porcupinetree · May 19, 2016

Re: MX1 Calculus Marathon 2016

Nailgun said:
wait you can't assume this lol?

si2136 said:
You can?

Pretty sure you can assume it. The proof is only a few lines anyway, it wouldn't hurt to chuck it in

si2136 · May 19, 2016

Re: MX1 Calculus Marathon 2016

porcupinetree said:
Pretty sure you can assume it. The proof is only a few lines anyway, it wouldn't hurt to chuck it in

I meant to say that you can assume it aah

leehuan · May 20, 2016

Re: MX1 Calculus Marathon 2016

porcupinetree said:
Pretty sure you can assume it. The proof is only a few lines anyway, it wouldn't hurt to chuck it in

si2136 said:
I meant to say that you can assume it aah

Never recommended as whilst the old syllabus still has it, it is never explicitly taught.

In the HSC: Remember the proof using e^(ln.c) = c to derive it. Like porcupinetree stated, its basically a 2 line proof.

Fortunately, it is not really examined these days either. This means its not a massive worry

HSC 2016 MX1 Marathon - Harder 3U Integration (archive) (1 Viewer)

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