The Student Room Group

help differentiating a quotient

Question:

given

y=4ln(x3)4ln(x+3)y=\dfrac{4\ln(x-3)}{4\ln(x+3)}

show that

dydx=24x(4ln(x+3))\dfrac{dy}{dx}=\dfrac{24}{x(4\ln(x+3))}

__________________________-



I can't seem to get there. Cancelling the 4's and using the quotient rule gives me

dydx=1(x3)ln(x+3)ln(x3)(x+3)(ln(x+3))2\dfrac{dy}{dx}=\dfrac{1}{(x-3)\ln(x+3)}-\dfrac{\ln(x-3)}{(x+3)(\ln(x+3))^2}

which doesn't look much like what they've got
(edited 12 years ago)
I get the same answer as you, what do they have?
Original post by jameswhughes
I get the same answer as you, what do they have?


they have what I posted. "show that"
Original post by Plato's Trousers
they have what I posted. "show that"


oops, didn't see that. i can't get it either :/
Reply 4
They've made a mistake I think, try plugging in values
Reply 5
try using the product rule instead.
Reply 6
I haven't used the qoutient rule in a few weeks, i guess il have to go back to it and recap.
Ewan has already sorted this one.
Reply 8
Use implicit differentiation, works for me
Original post by Scottishsiv
Use implicit differentiation, works for me


Could you explain please? :smile:
Original post by Scottishsiv
Use implicit differentiation, works for me

There is nothing implicit to differentiate here - everything's explicit unless you're trying to turn something that is explicit into an implicit form, but that almost never helps.
(edited 12 years ago)
Original post by Farhan.Hanif93
There is nothing implicit to differentiate here, everything's explicit.


I think he meant you could multiply both sides by the denominator.

However substitution shows the answer Plato has got (which is correct) is not equivalent to the supposedly simplified one anyway.
Well first turn it into:

yln(x+3) = ln(x-3)

then differentiate both sides using implicit differentiation

the you get dy/dx ln(x+3) = 1/(x-3) - y/(x+3)

then sub in y = ln(x-3) / ln (x+3)

I think that works, haven't worked further than this though. apologies for the layout.
Original post by Mr M
I think he meant you could multiply both sides by the denominator.

However substitution shows the answer Plato has got (which is correct) is not equivalent to the supposedly simplified one anyway.

Yep, edited my post just as you were typing.
Original post by Scottishsiv
Well first turn it into:

yln(x+3) = ln(x-3)

then differentiate both sides using implicit differentiation

the you get dy/dx ln(x+3) = 1/(x-3) - y/(x+3)

then sub in y = ln(x-3) / ln (x+3)

I think that works, haven't worked further than this though. apologies for the layout.

From which it follows that you get what Plato got, which is the right answer but it doesn't match the given answer.
thanks everyone! Reassuring :biggrin:

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