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c4 help.. again

how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
Original post by hajs
how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??


You can ignore the modulus within the ln when differentiating. It is introduced during integration.
Original post by hajs
how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??


x^2 + 1 is always positive, so don't worry too much about the modulus sign.

set, u = x^2 + 1, then y = ln u.

Plug into the chain rule.
Original post by hajs
how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??


Differentiate lnu where u = x^2 + 1

Becomes [1/u]*(du/dx) iirc
(edited 9 years ago)
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hajs
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Original post by morgan8002
You can ignore the modulus within the ln when differentiating. It is introduced during integration.


no way! I did not know that! haha :colondollar: okay so that would be 2x right?? so why does the book say y = ln|x^2 +1| dy/dx = 1/x^2+1 * 2x ??
nevermind this......................
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davros
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Original post by hajs
how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??


If x is real then |x^2 +1| = x^2 + 1 so you can just ignore the modulus :smile:

Just apply the standard chain rule with y = ln u where u = x^2 + 1.
Original post by Mr Inquisitive
Differentiate lnu where u = x^2 + 1

Becomes [1/(du/dx)]*u iirc


Actually it's dudx1u\frac{du}{dx}\frac{1}{u}
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hajs
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Original post by Lolgarithms
x^2 + 1 is always positive, so don't worry too much about the modulus sign.

set, u = x^2 + 1, then y = ln u.

Plug into the chain rule.


OHHHHH. oh yeah my bad haha. thank you!!!! :biggrin::biggrin::biggrin:
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Hasufel
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ddx(ln(f(x))=f′(x)f(x) \displaystyle \frac{d}{dx} (ln(f(x))= \frac{f ' (x)}{f(x)}

the "mod" doesn`t matter: x^2+1 is always +ve..
Original post by morgan8002
Actually it's dudx1u\frac{du}{dx}\frac{1}{u}


That's the one... Been two years
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hajs
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Original post by morgan8002
Actually it's dudx1u\frac{du}{dx}\frac{1}{u}


Original post by davros
If x is real then |x^2 +1| = x^2 + 1 so you can just ignore the modulus :smile:

Just apply the standard chain rule with y = ln u where u = x^2 + 1.


thanks i get it!!
Original post by hajs
no way! I did not know that! haha :colondollar: okay so that would be 2x right?? so why does the book say y = ln|x^2 +1| dy/dx = 1/x^2+1 * 2x ??


You can ignore the modulus, but use the substitution u = x^2 + 1.

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