Just wondering if anyone would assist me with some calculus?

13 Mar 2010 | 03:00 am | Category: portal.qdbs.com      edit
  • I've got a calculus test coming up and I just need some refreshing on some things. I understand a lot of the stuff, but there is a small portion of it that just doesn't seem to make sense to me. Or maybe I'm just looking at it the wrong way.

    1. Find y' if [x^(3y) = y^(8x)]

    2. Find dy/dx by implicit differentiation. 2sinx + 7cosy = sinxcosy

    3. Differentiate the function. Simplify. f(t)=[(6+lnt)/(4-lnt)]

    As for right now I think that's all I need assistance with. Any help is appreciated.


  • Good questions.
    If this is all the assistance you need right now, it looks like you're doing fine.

    1) Differentiate the left side with respect to x (treating y as if it was a constant) gives on the left: (3y)x^(3y - 1) dx/dx = (3y)x^(3y - 1).
    Remembering that d (e^x) / dx = e^x and that ln(e^z) = z, we can rewrite the right side as: y^(8x) = ln{e^[y^(8x)]}. Differentiate the right side with respect to x: (1 / {e^[y^(8x)]}) d {e^[y^(8x)]} / dx and d {e^[y^(8x)]} / dx = e^[y^(8x)] [(8x)y^(8x - 1)] dy/dx.
    Now, solve for dy/dx.

    2) 2sinx + 7cosy = (sinx)(cosy)
    So, 2cosx dx/dx - 7siny dy/dx =
    (sinx)(- siny dy/dx) + (cosy)(cosx dx/dx).
    2cosx - (cosy)(cosx)
    = [7siny - (sinx)(siny) ] dy/dx and
    dy/dx = cosx[2 - cosy] / siny[7 - sinx]

    3) Just use the Quotient Rule that you have memorized and that the derivative of
    ln[ f(w) ] with respect to w is
    [1 / f(w)] (df(w) / dw).







    • #If you have any other info about this subject , Please add it free.#
    Your name:
    E-mail:
    Telphone:

    Your comments:


    If you have any other info about Just wondering if anyone would assist me with some calculus? , Please add it free.
    edit