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Math Help - [SOLVED] Help with derivatives!

  1. #1
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    [SOLVED] Help with derivatives!

    I am trying to figure this thing out. I am supposed to get the derivative using the product rule, the equation is d/dx * (f(x)^4). I have gotten it down to f(x) * d/dx * (f(x)^3). I am completely stuck. Could anyone help point me in the right direction?
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  2. #2
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    I'm going to be a little picky here because I think it will help with your understanding. You don't really have an equation here, as an equation needs what? That's right, an equal sign, =.

    The terminology d/dx is meant to indicate that the next term is to be differentiated with respect to x.

    Do you have written f'(x)*f^4(x) OR \frac{d}{dx} f^4(x)?
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  3. #3
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    The 2nd equation is what I have written. I think it is more of a proof than an equation. Please, be as picky as you would like. It will help me to learn it better.
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  4. #4
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    Ok good I'm glad you see it's only to help.

    So if we have some something f(x) raised to the fourth power, we know that this is a sum of terms of various powers of x, meaning, the standard power rule can apply.

    So...

    \frac{d}{dx} f^4(x)=4f^3(x)*f'(x)

    Reduce the power by one and multiply by the derivative of the inner expression.

    Got it?
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  5. #5
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    I think iv'e got it. So if I had d/dx*f^5x, would it equal 5f^4x * f ' (x)?
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  6. #6
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    i thought the product rule was simply


    u * dv + v * du

    ^ if this is right you take your two terms( the terms from the problem in the book). make one of them u and the other v.

    so u multiply the u times the derivative of the v. then u add it to v times the derivative of u. simple as that *i think*


    please post the orignal problem.
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  7. #7
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    Quote Originally Posted by UNTEng10 View Post
    I think iv'e got it. So if I had d/dx*f^5x, would it equal 5f^4x * f ' (x)?
    Yes.

    Don't write the \frac{d}{dx} TIMES an expression because you're not multiplying them, you're taking d/dx OF an expression.
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  8. #8
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    Quote Originally Posted by Legendsn3verdie View Post
    i thought the product rule was simply


    u * dv + v * du

    ^ if this is right you take your two terms( the terms from the problem in the book). make one of them u and the other v.

    so u multiply the u times the derivative of the v. then u add it to v times the derivative of u. simple as that *i think*


    please post the orignal problem.
    Yes you are right this is the product rule. I realize the OP said to use it but I don't think it's the best way.

    Imagine f(x)=1+x+x^2.

    Thus (f(x))^4=(1+x+x^2)^4 and the way I showed holds as the way to calculate the derivative. Do you see that I can keep adding any degree of x and this still holds?
    Last edited by Jameson; September 22nd 2008 at 09:27 PM. Reason: different notation to help
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  9. #9
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    Quote Originally Posted by Jameson View Post
    Yes.

    Don't write the \frac{d}{dx} TIMES an expression because you're not multiplying them, you're taking d/dx OF an expression.
    Awesome, thanks for all the help!
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  10. #10
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    Quote Originally Posted by Jameson View Post
    Yes you are right this is the product rule. I realize the OP said to use it but I don't think it's the best way.

    Imagine f(x)=1+x+x^2.

    Thus f^4(x)=(1+x+x^2)^4 and the way I showed holds as the way to calculate the derivative. Do you see that I can keep adding any degree of x and this still holds?

    i kinda see but x^2+x+1 doesnt factor hmm.
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  11. #11
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    The chain rule says basically that if you have two functions, f(x) and g(x), then [f(g(x))]' = f'(g(x))*g'(x) So let g(x) just be some function g(x) and let f(x)=x^4. Now f(g(x)) = g^4(x) Right?

    So f' by itself is obviously 4x^3, now just let x be g(x) and we get the composite derivative is [4g(x)]^3*g'(x)

    Make sense?
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  12. #12
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    Quote Originally Posted by Jameson View Post
    The chain rule says basically that if you have two functions, f(x) and g(x), then [f(g(x))]' = f'(g(x))*g'(x) So let g(x) just be some function g(x) and let f(x)=x^4. Now f(g(x)) = g^4(x) Right?

    So f' by itself is obviously 4x^3, now just let x be g(x) and we get the composite derivative is [4g(x)]^3*g'(x)

    Make sense?

    yes, the chain rule, the stuff i forgot about the chain rule but now i remember! thanks!
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  13. #13
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    Thanks for the help guys, made my life a lot easier! Thanks sent out!
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