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# c4 help.. again watch

1. how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
2. (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.
3. (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.
4. (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
5. (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 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 ??
6. nevermind this......................
7. (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

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

Becomes [1/(du/dx)]*u iirc
Actually it's
9. (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!!!!

10. the "mod" doesn`t matter: x^2+1 is always +ve..
11. (Original post by morgan8002)
Actually it's
That's the one... Been two years
12. (Original post by morgan8002)
Actually it's
(Original post by davros)
If x is real then |x^2 +1| = x^2 + 1 so you can just ignore the modulus

Just apply the standard chain rule with y = ln u where u = x^2 + 1.
thanks i get it!!
13. (Original post by hajs)
no way! I did not know that! haha 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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