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Math Help - Differentiate questions

  1. #1
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    Differentiate questions

    Hi
    The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
    1) y=sin^{-1}(-2x-1)
    \frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}
    =\frac{-2}{\sqrt{1-(4x^2-1)}}
    =\frac{-2}{\sqrt{2-4x^2}}

    Answers says its: \frac{1}{\sqrt{x-x^2}}

    2) y=sin^{-1}(\sqrt{x})
    \frac{dy}{dx}= \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})
    =\frac{1}{2\sqrt{x}(\sqrt{1-x^2})}

    Answers says its: \frac{1}{2\sqrt{x-x^2}}

    3)Find the second derivative of xe^{2x}.
    The first derivative is e^{2x}(1+2x)
    The second derivative i got:
    =2e^{2x}+4xe^{2x}
    =2e^{2x}(1+2x)
    Answers says its 4(x+1)e^{2x}

    4)If y=2cos3x+5sin3x, show that  \frac{d^2y}{dx^2}=-9y
    First derivative i got: -6sin3x+15cos3x
    Now second derivative i got: -18cos3x-45sin3x
    What did i do wrong? or what is the next step?

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
    1) y=sin^{-1}(-2x-1)
    \frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}

    \textcolor{red}{=\frac{-2}{\sqrt{1-(-2x-1)^2}}}

    Answers says its: \frac{1}{\sqrt{x-x^2}}

    2) y=sin^{-1}(\sqrt{x})
    \frac{dy}{dx}= \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})

    \textcolor{red}{\frac{1}{2\sqrt{x} \cdot \sqrt{1-x}}}

    Answers says its: \frac{1}{2\sqrt{x-x^2}}

    3)Find the second derivative of xe^{2x}.
    The first derivative is e^{2x}(1+2x)
    The second derivative i got:
    =2e^{2x}+4xe^{2x}

    \textcolor{red}{=2e^{2x}+(1+2x)2e^{2x}}

    \textcolor{red}{=2e^{2x}+2e^{2x}+4xe^{2x}}

    \textcolor{red}{=4e^{2x}+4xe^{2x}}

    \textcolor{red}{4e^{2x}(1+x)}

    Answers says its 4(x+1)e^{2x}

    4)If y=2cos3x+5sin3x, show that  \frac{d^2y}{dx^2}=-9y
    First derivative i got: -6sin3x+15cos3x
    Now second derivative i got: -18cos3x-45sin3x
    What did i do wrong? or what is the next step? factor out a -9

    P.S
    ...
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  3. #3
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    3)Find the second derivative of
    The first derivative is
    The second derivative i got:


    <<<<<<<<<<<<<<







    can you explain to me how did you go from 2e^{2x}+4xe^{2x} to 2e^{2x}+(1+2x)2e^{2x}?
    Last edited by Paymemoney; February 11th 2010 at 04:26 PM.
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    3)Find the second derivative of
    The first derivative is
    The second derivative i got:


    <<<<<<<<<<<<<<







    can you explain to me how did you go from 2e^{2x}+4xe^{2x} to 2e^{2x}+(1+2x)2e^{2x}?
    I did not use your second derivative.

    y' = e^{2x}(1+2x) is where I started.

    use the product and chain rules from here as i did.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    I did not use your second derivative.

    y' = e^{2x}(1+2x) is where I started.

    use the product and chain rules from here as i did.
    oh ok i get it now
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  6. #6
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    So from \frac{-2}{\sqrt{1-(-2x-1)^2}} this is what i done but it doesn't seem to be correct.
    <br />
=\frac{-2}{\sqrt{1-(-2x-1)(-2x-1)}}

    =\frac{-2}{\sqrt{1-(4x^2+4x+1)}}

    =\frac{-2}{\sqrt{-4x^2+4x}}

    =\frac{-2}{\sqrt{-4x^2+4x}}

    =\frac{-2}{-2x^2+2x}

    =\frac{1}{x^2+x}
    Last edited by Paymemoney; February 11th 2010 at 06:05 PM.
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  7. #7
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    looking back at the original problem ...

    is the function y = \sin^{-1}(2x-1) or y = \sin^{-1}(-2x-1) as you originally posted ???

    the given solution is for the first function.
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  8. #8
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    Quote Originally Posted by skeeter View Post
    looking back at the original problem ...

    is the function y = \sin^{-1}(2x-1) or y = \sin^{-1}(-2x-1) as you originally posted ???

    the given solution is for the first function.
    its the first one y = \sin^{-1}(2x-1)
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  9. #9
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    Quote Originally Posted by Paymemoney View Post
    its the first one y = \sin^{-1}(2x-1)
    then work it again. you need the algebra practice.
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  10. #10
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    well i worked it out thanks for helping.
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  11. #11
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    Quote Originally Posted by Paymemoney View Post
    Hi
    The following questions i am having problem getting the correct answers. I need help where i have gone wrong.
    1) y=sin^{-1}(-2x-1)
    \frac{dy}{dx}=\frac{-2}{\sqrt{1-(2x-1)^2}}
    =\frac{-2}{\sqrt{1-(4x^2-1)}}
    The square of 2x- 1 is NOT 4x^2- 1!

    =\frac{-2}{\sqrt{2-4x^2}}

    Answers says its: \frac{1}{\sqrt{x-x^2}}

    2) y=sin^{-1}(\sqrt{x})
    \frac{dy}{dx}= \frac{1}{2\sqrt{x}} divide (\sqrt{1-x^2})
    =\frac{1}{2\sqrt{x}(\sqrt{1-x^2})}

    Answers says its: \frac{1}{2\sqrt{x-x^2}}

    3)Find the second derivative of xe^{2x}.
    The first derivative is e^{2x}(1+2x)
    The second derivative i got:
    =2e^{2x}+4xe^{2x}
    =2e^{2x}(1+2x)
    Answers says its 4(x+1)e^{2x}

    4)If y=2cos3x+5sin3x, show that  \frac{d^2y}{dx^2}=-9y
    First derivative i got: -6sin3x+15cos3x
    Now second derivative i got: -18cos3x-45sin3x
    What did i do wrong? or what is the next step?

    P.S
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