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Math Help - Derivatives

  1. #1
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    Derivatives

    Hi!

    I just started learning calculus this semester and it's a total brainbreak. We are on derivatives now and we just had a quiz about derivatives and limits. The problem is, when I thought I already knew derivatives I felt stupid during the whole 3 hours of answering the quiz. I found out I couldn't work on the complicated equations (especially the ones with the bar lines and radical signs) using the method our instructor taught us. I ended up having answered only one item which was a limit problem. Then a senior told me that I should have used the theorems for derivatives (not sure how to call that).

    Here's one from our quiz:
    y=(1+3x-x^2)(x^2-5)

    \frac {dy}{dx} = (x^2-5) \frac {d}{dx} (1+3x-x^2) + (1+3x-x^2) \frac {d}{dx} (x^2-5)

    \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) +  \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]

    \frac {dy}{dx} = (x^2-5) (3+2x) + (1+3x-x^2) (2x)

    \frac {dy}{dx} = 2x^3 + 3x^2 - 10x - 15 - 2x^3 + 6x^2 +2x

    \frac {dy}{dx} = 9x^2 - 8x - 15
    Is my answer correct? Unfortunately I can no longer get any mark for getting it right so it's good as incorrect.

    Thanks in advance!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    that's the way its done except in your first term it should be (x^2-5)*(3-2x) not 3+2x
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  3. #3
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    Quote Originally Posted by Xerxes View Post

    Here's one from our quiz:
    y=(1+3x-x^2)(x^2-5)

    \frac {dy}{dx} = (x^2-5) \frac {d}{dx} (1+3x-x^2) + (1+3x-x^2) \frac {d}{dx} (x^2-5)

    \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x) +  \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac {d}{dx}(5) ]

    \frac {dy}{dx} = (x^2-5) (3+2x) + (1+3x-x^2) (2x)
    In those lines there's something wrong, figure out by yourself.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    In those lines there's something wrong, figure out by yourself.
    Oh! Is there?
    Okay, maybe I'll check it by morning. I have to get some rest now. Thanks a lot.
    By the way, is there anything wrong about my syntax?
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    I see nothing wrong in your syntax.
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  6. #6
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    Talking crap!

    Quote Originally Posted by Calculus26 View Post
    that's the way its done except in your first term it should be (x^2-5)*(3-2x) not 3+2x
    Oh no! I did it again!
    That's not fair. I can't screw up in the exams because of signs.

    \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x)<br />
- \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac<br />
{d}{dx}(5) ]

    \frac {dy}{dx} = (x^2-5) (3-2x) + (1+3x-x^2) (2x)

    \frac {dy}{dx} = 3x^2 - 2x^3 + 10x - 15 - 2x^3 + 6x^2 +2x

    \frac {dy}{dx} = 9x^2 + 12x - 15 - 4x^3

    Now, how do I thank you two, electronically? (EDIT: I see. I just have to click the right button.)
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  7. #7
    MHF Contributor

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    Quote Originally Posted by Xerxes View Post
    Oh no! I did it again!
    That's not fair. I can't screw up in the exams because of signs.

    \frac {dy}{dx} = (x^2-5) [ \frac {d}{dx} (1) + \frac {d}{dx}(3x)<br />
- \frac {d}{dx} (x^2) ] + (1+3x-x^2) [ \frac {d}{dx} (x^2) - \frac<br />
{d}{dx}(5) ]

    \frac {dy}{dx} = (x^2-5) (3-2x) + (1+3x-x^2) (2x)

    \frac {dy}{dx} = 3x^2 - 2x^3 + 10x - 15 - 2x^3 + 6x^2 +2x

    \frac {dy}{dx} = 9x^2 + 12x - 15 - 4x^3

    Now, how do I thank you two, electronically? (EDIT: I see. I just have to click the right button.)
    Or slide twenty dollar bills into that little slot under your monitor!
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