Interesting possible trick (1 Viewer)

[Slidey]

If you were asked to differentiate something like

y=x(x^2-2x)^5/(x^3+1)^7

You would probably attempt to use the quotient rule, right?

But maybe it would be wise to try some implicit differentiation instead?

lny=lnx+5ln(x^2-2x)-7ln(x^3+1)
y'/y=1/x+10(x-1)/(x^2-2x)-21x^2/(x^3+1)
y'=y(1/x+10(x-1)/(x^2-2x)-21x^2/(x^3+1))

.'. y'=x(x^2-2x)^5/(x^3+1)^7(1/x+10(x-1)/(x^2-2x)-21x^2/(x^3+1))

Can anybody see any flaws with this method?

Let's test for y=tanx
y=sinx/cosx
y' is known to be sec^2(x)

lny=ln(sinx)-ln(cosx)
y'/y=cosx/sinx+sinx/cosx
y'/y=(cos^2(x)+sin^2(x))/(sinxcosx)
y'/y=1/(sinxcosx)
y'=y/(sinxcosx)=sinx/cosx*1/(sinxcosx)=1/cos^2(x)=sec^2(x)

So, yeah, it works. Obviously in cases where the top and the bottom are not functions composed of multiples, the quotient rule may be quicker. Otherwise I imagine this way would be.

 

[acmilan]

Nothing wrong with it, I like to use that method especially for double variable differentiation.

 

[KFunk]

I like it, thanks slide.

 

[dawso]

yeah steele, looks good, actually looks so obvious im amazed that it isnt well known, if i use it in my trial ill be sure 2 state "by steele's method"

 

[FinalFantasy]

that's cool, good work captain!

 

[acmilan]

Captain Slide does it again

 

[shafqat]

Yes, it's a nice trick. If anyone has Cambridge year 11, see ex. 12B extension for some questions using it. Here are are a few:
Differentiate: sqrt(x)sqrt(x+1)sqrt(x+2)
sqrt(x)(x-1)^2/(x+1).
x^ (1/x)
The various log laws do come in handy.

 

[lfc_reds2003]

OI Slidey that is a chilla method

were you just working sum sums and it came to you or were you taught it..

 

[nit]

It's a pretty standard little trick - it's done in Cambridge yr 11, as Shafqat said, so it's well known.

 

[Slidey]

aural: Read about it a while ago in some uni textbook. Forgot the method, but remembered that to prove for y=e^x, y'=e^x, you go:
y=e^x, lny=x, y'/y=1, y'=y, y'=e^x, so I tried using lny.

Lesse, shafqat...
y=sqrt(x)sqrt(x+1)sqrt(x+2)
lny=0.5(lnx+ln(x+1)+ln(x+2)
y'/y=0.5(1/x+1/(x+1)+1/(x+2))
y'=sqrt(x)sqrt(x+1)sqrt(x+2)[1/x+1/(x+1)+1/(x+2)]/2

Neat. So it's good for more than just replacing quotient rule.

y=sqrt(x)(x-1)^2/(x+1).
lny=1/2ln(x)+2ln(x-1)-ln(x+1)
y'/y=1/(2x)+2/(x-1)-1/(x+1)
y'=sqrt(x)(x-1)^2[1/(2x)+2/(x-1)-1/(x+1))/(x+1]

y=x^(1/x)
lny=ln(x)/x
y'/y=1-ln(x)/x^2
y'=x^(1/x)-x^([1-2x]/x)ln(x)

Doesn't give elegant solutions... but then, I suppose neither would the product or quotient rules, and it's a hell of a lot faster.

 

[SeDaTeD]

Would that work for all functions, or only those which are always positive, since you can't log a negative number? (unless this extends to complex values, which seems reasonable enough).
Hmm, I think it probably does work, might need to look up somewhere for a justification for all functions i suppose.

 

[dawso]

it will never not work, as in, give an incorrect answer, it may however be long or messy or not workable in which case u go bak 2 the oldschool techniques

 

[casebash]

Yep, us in Maths 1 at Ruse were taught this in class.

 

[dawso]

of course u were.....well i learnt it off steele, and man, in my 4unit trial yesterday...i used it, i had to find the stationary points of a bitchy graph and i used it and said "by steele's" technique, u shoudl be proud man...

 

[Slidey]

Hahaha! Thanks, dawso.

 

[justchillin]

Nice method, I have seen this done in a random trial before, I think western region where they asked you to do exactly the same thing using f(x) and g(x), etc...

 

[stryder_au]

this is gold. thanks slide!