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Core 3 Differentiation watch

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    How do I differentiate 2ln(x/4)? The answer is 2/x but I am not sure what the steps are to get that answer
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    (Original post by Yusufff)
    How do I differentiate 2ln(x/4)? The answer is x/2 but I am not sure what the steps are to get that answer
    dont you mean  \displaystyle \frac{2}{x}?

    well as a generall rule,  \displaystyle \frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} if that helps you any
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    (Original post by DylanJ42)
    dont you mean  \displaystyle \frac{2}{x}?

    well as a generall rule,  \displaystyle \frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} if that helps you any
    Yeah sorry typo Thanks!
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    (Original post by DylanJ42)
    dont you mean  \displaystyle \frac{2}{x}?

    well as a generall rule,  \displaystyle \frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} if that helps you any
    I have another problem, probably a silly mistake but:
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    How did the answer get a 1 on the numerator?


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    What is  \frac{dy}{dx}, y = \frac{x-1}{2}? That's where you're going wrong.
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    (Original post by trythis)
    What is  \frac{dy}{dx}, y = \frac{x-1}{2}? That's where you're going wrong.
     y = \frac{x-1}{2},  \frac{dy}{dx} = \frac{1-1}{2} = \frac{0}{2}

    This is what I'm getting, I can't spot the silly mistake o-o
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    y = \frac {1}{2}x - \frac{1}{2}

    Try differentiating that, remembering that differentiating a constant gives you zero
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    (Original post by drinktheoceans)
    y = \frac {1}{2}x - \frac{1}{2}

    Try differentiating that, remembering that differentiating a constant gives you zero
    That just leaves you with 1/2 right? ...Aaand that's what I was doing wrong:facepalm: TY!
 
 
 
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