Lovely ln

Calculus Level 1

d d x ln ∣ x ∣ = ? \Large \dfrac{d}{dx} \text{ln}|x| = \ ?

Note: Here x ≠ 0 x \neq 0 .

− 1 x -\dfrac{1}{x} 1 x \dfrac{1}{x} 0 Not defined

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Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Rishabh Jain
Jan 18, 2016

1. x > 0 ( ∣ x ∣ = x ) \color{magenta}{1.x>0}\quad\quad(|x|=x) d d x ln x = 1 x \dfrac{d}{dx} \text{ln}x=\dfrac{1}{x} 2. x < 0 ( ∣ x ∣ = − x ) \color{magenta}{2.x<0}\quad\quad(|x|=-x) d d x ln ( − x ) = 1 ( − x ) × ( − 1 ) = 1 x ( Chain Rule ) \dfrac{d}{dx} \text{ln}(-x)=\dfrac{1}{(-x)}\times(-1)=\dfrac{1}{x}\quad\quad (\color{#20A900}{\text{Chain Rule}})

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Cool , isn't it?

Nihar Mahajan - 5 years, 4 months ago

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Yup..... :)

Rishabh Jain - 5 years, 4 months ago
Akhil Bansal
Jan 19, 2016

d d x ln ⁡ ∣ x ∣ = 1 ∣ x ∣ d d x ∣ x ∣ = 1 ∣ x ∣ ⋅ ∣ x ∣ x = 1 x \large \dfrac{d}{dx} \ln |x| = \dfrac{1}{|x|} \dfrac{d}{dx} |x| = \dfrac{1}{|x|} \cdot \dfrac{|x|}{x} = \dfrac{1}{x}

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Moderator note:

Be careful with the first equation. While it is factually true, you haven't explained why we can pull out the ∣ x ∣ |x| in such a way.

Be careful with the first equation. While it is factually true, you haven't explained why we can pull out the ∣ x ∣ |x| in such a way.

Calvin Lin Staff - 5 years, 4 months ago

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