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    Hi, I was wondering how to differentiate cube root 4x^4. Ive tried using the chain rule, but I just cant seem to get the answer?
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    (Original post by RNVS_99)
    Hi, I was wondering how to differentiate cube root 4x^4. Ive tried using the chain rule, but I just cant seem to get the answer?
    How did you apply the chain rule? Please show working.
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    (Original post by RNVS_99)
    Hi, I was wondering how to differentiate cube root 4x^4. Ive tried using the chain rule, but I just cant seem to get the answer?
    \sqrt[3]{4x^4} = 4^{1/3}\cdot x^{4/3}

    Why are you even using the chain rule...?
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    (Original post by RNVS_99)
    Hi, I was wondering how to differentiate cube root 4x^4. Ive tried using the chain rule, but I just cant seem to get the answer?
    Hint: \sqrt[3]{4x^4} = (4x^4)^\frac{1}{3}.

    Do you see where to go from there?
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    (Original post by RDKGames)
    \sqrt[3]{4x^4} = 4^{1/3}\cdot x^{4/3}

    Why are you even using the chain rule...?
    Its a function within a function, so I thought the chain rule would be appropriate, but Im not to sure anyway?
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    (Original post by RNVS_99)
    Its a function within a function, so I thought the chain rule would be appropriate, but Im not to sure anyway?
    You can use the chain rule, sure, but since there is no constant inside this cube root then you can do this instead as it's faster since you can split the terms.
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    (Original post by marioman)
    Hint: \sqrt[3]{4x^4} = (4x^4)^\frac{1}{3}.

    Do you see where to go from there?
    Thats exactly what I did, but I cant get the answer of 4/3 cube root 4x
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    dy/dx = nx^n-1
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    (Original post by RDKGames)
    You can use the chain rule, sure, but since there is no constant inside this cube root then you can do this instead as it's faster since you can split the terms.
    I still cant manage to get the answer of 4/3 cube root 4x
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    (Original post by RNVS_99)
    I still cant manage to get the answer of 4/3 cube root 4x
    That's not the full answer. Show us your working out.
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    (Original post by RDKGames)
    That's not the full answer. Show us your working out.
    The answer is 4/3 cube root 4x as it says in the book
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    (Original post by RDKGames)
    That's not the full answer. Show us your working out.
    I dont know how to get the answer, which is why I am asking
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    (Original post by RNVS_99)
    I dont know how to get the answer, which is why I am asking
    If you're going with the form I've demonstrated then know that:

    \frac{d}{dx}(ax^n)=a \cdot n \cdot x^{n-1}

    Show us you applying this to the question.
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    (Original post by RDKGames)
    If you're going with the form I've demonstrated then know that:

    \frac{d}{dx}(ax^n)=a \cdot n \cdot x^{n-1}

    Show us you applying this to the question.
    So far, my answer is 4/3^-2/3 * 4x/3^1/3
    Am I right so far??
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    (Original post by RNVS_99)
    So far, my answer is 4/3^-2/3 * 4x/3^1/3
    Am I right so far??
    No. I'm not quite sure why you reduced the exponent on the 4...

    \frac{d}{dx}(4^{1/3} \cdot x^{4/3})=4^{1/3} \cdot \frac{4}{3} \cdot x^{1/3}
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    (Original post by RDKGames)
    No. I'm not quite sure why you reduced the exponent on the 4...

    \frac{d}{dx}(4^{1/3} \cdot x^{4/3})=4^{1/3} \cdot \frac{4}{3} \cdot x^{1/3}
    ahh okay, I see, cant you then multiply the 4 and the x, giving you the cube root 4x?
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    (Original post by RNVS_99)
    ahh okay, I see, cant you then multiply the 4 and the x, giving you the cube root 4x?
    Indeed. I've misread your previous answer, could help to use latex.
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    (Original post by RDKGames)
    Indeed. I've misread your previous answer, could help to use latex.
    haha, sorry about that, thanks for the help, this question confused me a bit, but now I understand it!
 
 
 
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