Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Logarithmic differentiation

Compute the derivative

\frac{d}{dx}\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]

Let

y=\frac{x^2+1}{x^3(x-1)^2}

Then using logarithmic differentiation let

\ln y=\ln\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]=\ln(x^2+1)-3\ln x-2\ln(x-1)

\frac{d}{dx}(\ln y) = \frac{d}{dx}\big[\ln(x^2+1)-3\ln x-2\ln(x-1)\big]

\frac{1}{y}\frac{dy}{dx} = \frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}

\frac{dy}{dx}=y\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big]

which equals

\frac{x^2+1}{x^3(x-1)}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big]

? I'm a bit lost now, did I miss something? Wolfram is giving me a different answer...

(edited 7 years ago)

Original post by AishaGirl

Compute the derivative

\frac{d}{dx}\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]

The expressions will be equivalent, you're answer is correct.

OP

Original post by B_9710

The expressions will be equivalent, you're answer is correct.

My answer? I've not even finished differentiating yet. I must be missing something because when I try to get rid of the square brackets it's wrong.

Original post by AishaGirl

My answer? I've not even finished differentiating yet. I must be missing something because when I try to get rid of the square brackets it's wrong.

You have finished the differentiation part - next it's just algebra. It's not wrong, I checked your answer and it's correct, it's must just be in a different form to the 'correct' answer that you saw.

OP

Original post by B_9710

You have finished the differentiation part - next it's just algebra. It's not wrong, I checked your answer and it's correct, it's must just be in a different form to the 'correct' answer that you saw.

Wolfram alpha says that it's

\frac{2}{x^3(x-1)^2}

for

\frac{x^2+1}{x^3(x-1)}

where is this coming from? Is it

x^3(x-1)^2

or

x^3(x-1)

?

I'm so confused. I've missed something and I think you have too?

OP

Oh I have!

It's

\frac{x^2+1}{x^3(x-1)^2}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big]

which gives

\frac{2}{x^2(x-1)^2}-\frac{3(x^2+1}{x^4(x-1)^2}-\frac{2(x^2+1)}{x^3(x-1)^3}

=\frac{2x^2(x-1)-3(x^2+1)(x-1)-2(x^2+1)x}{x^3(x-1)^3}

and finally

\frac{-3x^3+x^2-5x+3}{x^4(x-1)^3}

which now matches what wolfram alpha says.

I was missing a power.

Original post by AishaGirl

Oh I have!

It's

\frac{x^2+1}{x^3(x-1)^2}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big]

which gives

\frac{2}{x^2(x-1)^2}-\frac{3(x^2+1}{x^4(x-1)^2}-\frac{2(x^2+1)}{x^3(x-1)^3}

=\frac{2x^2(x-1)-3(x^2+1)(x-1)-2(x^2+1)x}{x^3(x-1)^3}

and finally

\frac{-3x^3+x^2-5x+3}{x^4(x-1)^3}

which now matches what wolfram alpha says.

I was missing a power.

Right. I did notice you forgot the power of 2 of that term but I forgot to mention it - it was of course all right apart from that as you said.

Is logarithmic differentiation useful in A level? Or is this uni?

Original post by thekidwhogames

Is logarithmic differentiation useful in A level? Or is this uni?

it's useful for A2 Maths in C3, C4 and DE for MEI maths, dk about other boards though

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Original post by thekidwhogames

Is logarithmic differentiation useful in A level? Or is this uni?

no need for the bump :tongue:

:tongue:

but yeah applying log to both sides can be useful at A-level to avoid getting lost in a mess of algebra

Original post by _gcx

no need for the bump :tongue:

:tongue:

but yeah applying log to both sides can be useful at A-level to avoid getting lost in a mess of algebra

Alright, thanks!

Original post by NishatM

it's useful for A2 Maths in C3, C4 and DE for MEI maths, dk about other boards though

Yeah sorry about that. I was watching a video by Prof Leonard and he went over this so it seemed to be really useful. Thanks for the help!

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