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Math Help - Another differential.

  1. #1
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    Another differential.

    y = \sqrt {e^{5x}+3x}

    = (e^{5x}+3x^2)^{\frac{1}{2}}

    y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5x+6x)

    Is that as far as i go , im not sure what to do now .
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  2. #2
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    Quote Originally Posted by el123 View Post
    y = \sqrt {e^{5x}+3x}

    = (e^{5x}+3x^2)^{\frac{1}{2}}

    y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5x+6x)

    y'= \frac{1}{2} (e^{5x}+3x^2)^{-\frac{1}{2}} (\textcolor{red}{5e^{5x}}+6x)
    correction

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  3. #3
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    Quote Originally Posted by el123 View Post
    y = \sqrt {e^{5x}+3x}

    = (e^{5x}+3x^2)^{\frac{1}{2}}

    y'= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5e^{5x}+6x)

    Is that as far as i go , im not sure what to do now .
    If it is the differential that you want, first start by changing your notation. Instead of y', use Liebnitz notation. IE \frac{dy}{dx}

    Then all you'd have too do is multiply both sides by dx. Like so

    dx\left(\frac{dy}{dx}\right)= \frac{1}{2} (e^{5x}+3x^2)^{\frac{-1}{2}} (5e^{5x}+6x)dx
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  4. #4
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    so if im asked to find the slope of that equation when x = 1 how do i do that?

    Also could you please explain where the correction came from.

    And i really appreciate all your help skeeter. You rock.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    so if im asked to find the slope of that equation when x = 1 how do i do that?
    Plug 1 into the derivative.

    Quote Originally Posted by el123 View Post
    Also could you please explain where the correction came from.
    Recal that \frac{d}{dx}[e^u]=e^uu'. in your problem \frac{d}{dx}[e^{5x}]=e^{5x}\cdot{5}

    Quote Originally Posted by el123 View Post
    And i really appreciate all your help skeeter. You rock.
    I rock too
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  6. #6
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    Quote Originally Posted by el123 View Post
    so if im asked to find the slope of that equation when x = 1 how do i do that?

    Also could you please explain where the correction came from.
    the derivative of e^{f(x)} is e^{f(x)} \cdot f'(x)

    so, the derivative of e^{5x} is e^{5x} \cdot 5 = 5e^{5x}

    to find the value of the derivative at x = 1, plug in 1 for x.


    btw ... what you are finding is a derivative, not a differential.
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  7. #7
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    ok now im confused , so i am being asked to 'differentiate'

    Can you please explain where i begin to 'differentiate' as opposed to finding the derivative?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by el123 View Post
    ok now im confused , so i am being asked to 'differentiate'

    Can you please explain where i begin to 'differentiate' as opposed to finding the derivative?
    Yeah, it's easy. The process of finding the derivative is called differentiation.

    So, you differentiated y to find the derivative y'.

    I thought you wanted the differential in my first post, so forget that ever happened. You will know what differentials are soon enough...
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