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Math Help - New member , eager to learn but i might need some help!

  1. #1
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    New member , eager to learn but i might need some help!

    Hi everyone as per title, Im doinf a new math course and struggle with some ideas and concepts. Im very grateful for a site like this, it seems very knowledgable!
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  2. #2
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    Re: New member , eager to learn but i might need some help!

    Hi. There are many willing and able to help here.

    State the problem exactly and completely.

    If you have done some work, show it so we can see where you are stuck.

    If you have an answer, but are unsure, show your answer PLUS your work so, IF you made a mistake, we can see where.

    Otherwise, ask for a hint to get started.
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  3. #3
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    Re: New member , eager to learn but i might need some help!

    DY/DX e^5x sec (x^3).

    I have to differentiate this. I belive I have to use the product rule then the chain rule.

    f'x = e^5x d/dx ((sec (x^3)) + (sec (x^3)) d/dx ((e^5x).

    g(u) = sec (u) where u = f(x) = (x^3)

    now g'(u) = sec x tan x and f'x = 3x^2

    Composite rule = k'(x) = g'(f(x)) f' (x)

    = g' (u) f'(x)

    = sec x tan x . 3x^2

    f'x = e^5x (sec x tan x 3x^2) + (sec (x^3)) 5e^5x

    Thats it. I find this quite difficult.
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  4. #4
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    Re: New member , eager to learn but i might need some help!

    Hi Me too new here and just joined the things.
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  5. #5
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    Re: New member , eager to learn but i might need some help!

    Quote Originally Posted by NED1 View Post
    DY/DX e^5x sec (x^3).

    I have to differentiate this. I belive I have to use the product rule then the chain rule.

    f'x = e^5x d/dx ((sec (x^3)) + (sec (x^3)) d/dx ((e^5x).

    g(u) = sec (u) where u = f(x) = (x^3)

    now g'(u) = sec x tan x and f'x = 3x^2

    Composite rule = k'(x) = g'(f(x)) f' (x)

    = g' (u) f'(x)

    = sec x tan x . 3x^2

    f'x = e^5x (sec x tan x 3x^2) + (sec (x^3)) 5e^5x

    Thats it. I find this quite difficult.
    close but not quite.

    $\dfrac d {dx} \sec(x^3) = 3x^2 \sec(x^3)\tan(x^3)$ not $3x^2 \sec(x)\tan(x)$ as you've written.
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  6. #6
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    Re: New member , eager to learn but i might need some help!

    Quote Originally Posted by romsek View Post
    close but not quite.

    $\dfrac d {dx} \sec(x^3) = 3x^2 \sec(x^3)\tan(x^3)$ not $3x^2 \sec(x)\tan(x)$ as you've written.
    thank you! otherwise it is correct?
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  7. #7
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    Re: New member , eager to learn but i might need some help!

    Quote Originally Posted by NED1 View Post
    thank you! otherwise it is correct?
    D[ e^(5x) sec (x^3),x] - Wolfram|Alpha

    I believe so
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  8. #8
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    Re: New member , eager to learn but i might need some help!

    Thanks very much.
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