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Math Help - More Differentiation Questions

  1. #1
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    More Differentiation Questions

    Hi
    A couple of problems i am having trouble with:

    1) 3\sqrt{x}(x^2+2x)
    2)Show that \frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    1) 3\sqrt{x}(x^2+2x)

    The product rule says

    For y=u\times v \Rightarrow y' = u'\times v+u\times v'

    In your case make u = 3\sqrt{x} and v = x^2+2x

    Now find u' and v'
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    A couple of problems i am having trouble with:

    2)Show that \frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}

    P.S
    \frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)

    in this case, let f(x)=\sqrt{x} and g(x)=x^2+-a^2
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  4. #4
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    Quote Originally Posted by pickslides View Post
    The product rule says

    For y=u\times v \Rightarrow y' = u'\times v+u\times v'

    In your case make u = 3\sqrt{x} and v = x^2+2x

    Now find u' and v'
    thanks got the right answer
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  5. #5
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    Quote Originally Posted by dedust View Post
    \frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)

    in this case, let f(x)=\sqrt{x} and g(x)=x^2+-a^2
    i can't seen to get it correct.
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  6. #6
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    f'(x)=\frac{1}{2\sqrt{x}}
    g'(x)=2x

    then
    f'(g(x)) \times g'(x) = . . .
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  7. #7
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    isn't g'(x) = 2x +- 2a

    This is how i approached it:
    =\frac{1}{2\sqrt(x^2)(2x)}
    =\frac{1}{2\sqrt{2x^3}}

    This is where i get stuck i i don't know what can be done.
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  8. #8
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    a is a constant, it doesn't depend on x, so the derivative is 0.
    now you should be able to calculate f'(g(x)) \times g'(x)
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  9. #9
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    oh ok yeh i get it now
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