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Math Help - Differentiate with respect to x

  1. #1
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    Differentiate with respect to x

    Y = \displaystyle \sqrt{(x^3+5)}
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  2. #2
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    e^(i*pi)'s Avatar
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    Use the chain rule: \dfrac{d}{dx} f[g(x)] = f'[g(x)] \cdot g'(x)

    In your case you have f(x) = \sqrt{g(x)} and g(x) = x^3+5
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  3. #3
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    \dfrac{d}{dx}\left(\sqrt{u(x)}\right)=\dfrac{1}{2\  sqrt{u(x)}}\cdot u'(x)


    Fernando Revilla

    Edited: Sorry, I didn't see e^(i*pi)'s post.
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  4. #4
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    You can also write this as (x^3+5)^{1/2} to use the "power law" and the chain rule.
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  5. #5
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    Is this the final answer?

    =\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}
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  6. #6
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    How do I expand on this?
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  7. #7
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    e^(i*pi)'s Avatar
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    Yes that's fine, I wouldn't bother expanding that though.
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  8. #8
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    Quote Originally Posted by HNCMATHS View Post
    Is this the final answer?

    =\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}
    Let's see!

    \displaystyle\ u=x^3+5\Rightarrow\frac{du}{dx}=3x^2


    From the Chain Rule

    \displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du  }{dx}=\frac{d}{du}u^{\frac{1}{2}}\;3x^2=\frac{1}{2  }u^{-\frac{1}{2}}\;3x^2=\frac{1}{2u^{\frac{1}{2}}}3x^2=  \frac{3x^2}{2\sqrt{x^3+5}}

    So you see, those fast methods can be quite useful as an alternative to messing with fractional indices!
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    You can also write this as (x^3+5)^{1/2} to use the "power law" and the chain rule.
    HI Halls of Ivy it is this method I would like you to expand on please, Thanks
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  10. #10
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    Quote Originally Posted by HNCMATHS View Post
    HI Halls of Ivy it is this method I would like you to expand on please, Thanks
    Have you been taught the chain rule? Your class notes and textbook will have many examples for you to follow and relate to the question you posted.
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