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Math Help - Check my working please.

  1. #1
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    Check my working please.

    Differentiate these.


    y = \frac{1}{\sqrt{1-5x^2}}

    =(1-5x^2)^\frac{-1}{2}

    =-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)

    =\frac{5x}{\sqrt{(x^2-5)}^3}

    Next

    y = x^5e^(1-3x)

    =5x^4e^(1-3x)+(-3)(x^5)

    =e^(1-3x)(5-3x)

    Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.



    Can you please check those for me??
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  2. #2
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    Quote Originally Posted by el123 View Post
    Differentiate these.


    y = \frac{1}{\sqrt{1-5x^2}}

    =(1-5x^2)^\frac{-1}{2}

    =-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)

    =\frac{5x}{\sqrt{(x^2-5)}^3}

    Next

    y = x^5e^(1-3x)

    =5x^4e^(1-3x)+(-3)(x^5)

    =e^(1-3x)(5-3x)

    Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.



    Can you please check those for me??
    for the first derivative, how does the quantity (1-5x^2) become (x^2-5) ?


    y = x^5 \cdot e^{1-3x}

    y' = x^5 \cdot -3e^{1-3x} + e^{1-3x} \cdot 5x^4

    y' = x^4e^{1-3x}(5 - 3x)<br />
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by el123 View Post
    Differentiate these.


    y = \frac{1}{\sqrt{1-5x^2}}

    =(1-5x^2)^\frac{-1}{2}

    =-\frac{1}{2} (1-5x^2)^\frac{-3}{2} (-10x)

    =\frac{5x}{\sqrt{(x^2-5)}^3}

    Next

    y = x^5e^(1-3x)

    = 5x^4e^(1-3x)+(-3)(x^5) to make the power in right form write e^{1-3x} not e^(1-3x)

    =e^(1-3x)(5-3x)

    Just so you know , in that second equation , the e is supposed to be to the power of 1-3x. But latex wont let me show it for some reason. So please keep that in mind.



    Can you please check those for me??
    ....
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  4. #4
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    is that first one more like this

    \frac{5x}{\sqrt{(1-5x^2)^3}}

    Does that make more sense?
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  5. #5
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    Re:Above , Can someone tell me if im correct?
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    Re:Above , Can someone tell me if im correct?
    You are correct. But let me ask you a question. Is (x^{1/2})^3=(x^3)^{1/2} ?
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  7. #7
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    no i dont think so . Whys that seņor?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    no i dont think so . Whys that seņor?
    Because they are equal! That means that you had it right the first time!
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  9. #9
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    so why did someone ask how i got the first answer?"
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    so why did someone ask how i got the first answer?"
    Um, because I'm an idiot. I misread your first post. I thought the distinction lay in the laws of exponents, but I can see now that the problem was your differentiating. Sorry for the confusion.

    Yeah, in the first one, you forgot to keep that x^2 with that 5 throughout the process. Your second try is right though.

    Sorry, dude.

    But at least we learned something about exponents huh?
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  11. #11
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    its ok , im slowly compiling a mental file of how everything works , i am missing some of the math fundamentals so i find myself screwing up on basic things. Thanks for helping though.
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    its ok , im slowly compiling a mental file of how everything works , i am missing some of the math fundamentals so i find myself screwing up on basic things. Thanks for helping though.
    You are a gentleman, indeed.

    Again, sorry for the trouble.
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