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    Differentiating ln(1/3x)


    Compare the two workings, the 1st one is correct, the 2nd one is wrong. but Why?

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    (Original post by SaadKaleem)
    Differentiating ln(1/3x)


    Compare the two workings, the 1st one is correct, the 2nd one is wrong. but Why?
    Spoiler:
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    \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{3x} \right )= \frac{1}{3}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{x} \right )= -\frac{1}{3x^{2}}
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    (Original post by aymanzayedmannan)
    \displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{3x} \right )= \frac{1}{3}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{x} \right )= -\frac{1}{3x^{2}}
    Alright, thank you... can be quite tempting to write it as (3x)^-1
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    (Original post by SaadKaleem)
    Alright, thank you... can be quite tempting to write it as (3x)^-1
    ax^{n} can be written as \left( \sqrt[n]{a}x \right )^{n} but \frac{\mathrm{d}\left ( ax^{n} \right ) }{\mathrm{d} x} \neq \left ( \sqrt[n]{a}x \right )^{n-1}.

    Remember than constants are unchanged when differentiating (or integrating) linear functions.
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    (Original post by aymanzayedmannan)
    ax^{n} can be written as \left( \sqrt[n]{a}x \right )^{n} but \frac{\mathrm{d}\left ( ax^{n} \right ) }{\mathrm{d} x} \neq \left ( \sqrt[n]{a}x \right )^{n-1}.

    Remember than constants are unchanged when differentiating (or integrating) linear functions.
    Thanks once again.
 
 
 
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