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how to solve this?

Since I've given up on reading useless university notes, I still read them and found that they don't help.

...

Er, so yeah how the **** do you even begin with q2 and 3
IMG_20140421_214029.jpg329.8KB
Original post by MSI_10
Since I've given up on reading useless university notes, I still read them and found that they don't help.

...

Er, so yeah how the **** do you even begin with q2 and 3


Use the chain rule.

Rewrite 3 as f(x)=e(ln⁡2)(tan⁡1x)f(x)=e^{(\ln 2) (\tan \frac{1}{x})}
Reply 2
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MSI_10
OP
15
Original post by Mr M
Use the chain rule.

Rewrite 3 as f(x)=e(ln⁡2)(tan⁡1x)f(x)=e^{(\ln 2) (\tan \frac{1}{x})}


I thought chain rule but then questioned what is u and what is v.

Is U= sin and v the rest of it
Reply 3
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TenOfThem
18
Original post by MSI_10
I thought chain rule but then questioned what is u and what is v.

Is U= sin and v the rest of it


Do you actually need to set things up in terms of u and v - that is how we teach it for very basic examples
Reply 4
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MSI_10
OP
15
Original post by TenOfThem
Do you actually need to set things up in terms of u and v - that is how we teach it for very basic examples


Can only use what we've been taught. Haven't been given an explanation nor lecture in uni for using chain rule without u and v
Reply 5
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TenOfThem
18
Original post by MSI_10
Can only use what we've been taught. Haven't been given an explanation nor lecture in uni for using chain rule without u and v


Did you always use u and v at school
Even when the questions were simple
Even when there were more than 2 functions


You have 5 functions here so you will need more letters
(edited 10 years ago)
Original post by MSI_10
Can only use what we've been taught. Haven't been given an explanation nor lecture in uni for using chain rule without u and v


If that's all you know you might like to practise some easy examples until you get the hang of it.

If you want to get on with this question do it one step at a time.

Start by differentiating y=tan⁡3xy=\tan^3 x

Then try y=cos⁡2(tan⁡3x)y=\cos^2 (\tan^3 x)

Then try y=sin⁡(cos⁡2(tan⁡3x)y=\sin (\cos^2 (\tan^3 x)
Reply 7
Avatar for CTArsenal
CTArsenal
8
Original post by Mr M
If that's all you know you might like to practise some easy examples until you get the hang of it.

If you want to get on with this question do it one step at a time.

Start by differentiating y=tan⁡3xy=\tan^3 x

Then try y=cos⁡2(tan⁡3x)y=\cos^2 (\tan^3 x)

Then try y=sin⁡(cos⁡2(tan⁡3x)y=\sin (\cos^2 (\tan^3 x)


(Letting y=f(x){y} = f(x))

For the second one I just did arcsin⁡y=cos⁡2(tan⁡3x)\arcsin y = \cos^2 (\tan^3 x), and I somehow managed to come up with an answer *which I hope is right* just by using implicit differentiation and differentiating with two variables in the chain rule, i.e. letting tan⁡3x=u\tan^3 x = u, and ∴arcsin⁡y=cos⁡2u\therefore \arcsin y = \cos^2 u.

Ended up getting:

dydx=−3sin⁡(2tan⁡3x)cos⁡(cos⁡2(tan⁡3x))tan⁡2xsec⁡2x\frac{dy}{dx} = -3\sin(2\tan^3 x)\cos(\cos^2 (\tan^3 x))\tan^2 x\sec^2 x

Similarly for the first and third questions I got:

1) dydx=2x2(1−x3)43(1+x3)23\frac{dy}{dx} = \dfrac{2x^2}{(1-x^3)^\frac{4}{3}(1+x^3)^\frac{2}{3}}

3)
Unparseable latex formula:

\frac{dy}{dx} = \dfrac{-\ln2\sec^2(\frac{1}{x})\(2^\tan(\frac{1}{x})}{x^2}}



Just wanted to test myself since I'll probably encounter this stuff at uni next year, haha.

Are these right?
(edited 10 years ago)
Original post by CTArsenal

Spoiler



They are correct. However, I'd put them in a spoiler. Also, the answer to the third is slightly ambiguous. Is everything after ln⁡\ln the argument of the natural logarithm? (I know it isn't, but parentheses get rid of such problems.

Spoiler

(edited 10 years ago)
Reply 9
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MSI_10
OP
15
Think I got the first part right but the second part is probably wrong.
IMG_20140422_125150.jpg481.9KB
Reply 10
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MSI_10
OP
15
Okay realised I was using product rule instead like an idiot.

Is this fine for the first 2 parts
IMG_20140422_131907.jpg447.4KB
Reply 11
Avatar for CTArsenal
CTArsenal
8
Original post by Khallil
They are correct. However, I'd put them in a spoiler. Also, the answer to the third is slightly ambiguous. Is everything after ln⁡\ln the argument of the natural logarithm? (I know it isn't, but parentheses get rid of such problems.

Spoiler



ah okay, thanks very much :smile:

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