Results 1 to 2 of 2

Math Help - Differentiation, please HELP!

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    2

    Differentiation, please HELP!

    1. The problem statement, all variables and given/known data

    1) If f(x)= sin^4x, then f '(pi/3)
    2) Given f(x) = x/tanx, find f '(3pi/4)
    3) If f(x) = sinxcosx, then f '(pi/6)
    4) Differentiate: f(x) = x^2 + 2tanx
    Question that I have answered but not sure if it's really the right answer:
    5) Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1

    2. Relevant equations

    Product Rule?
    • y= f x g
      F'g + g'f

    Chain Rule?
    • y=f / g
    • ((f'g) - (g'f)) / ((g^2))

    Slope?
    • y=mx+b



    3. The attempt at a solution

    1)
    f(x) = sin^4x then f '(pi/3)
    Derivative of sinx = cosx therefore...
    I'll assume that sin^4x has the derivative of cos^4x
    Now, plug in the number...
    cos^4(pi/3) = 0.065

    Is that right?

    2)
    Given f(x) = x/tanx, find f '(3pi/4)
    Chain Rule:
    ((f'g) - (g'f)) / (g^2)
    Therefore...
    ((1 * tanX) - (??? * X)) / ((tanx^2))

    Let * be multiplication sign and the ??? to be the "I don't know".
    So, I got stuck of what the derivative of "tanx". However, what I do know is that:
    tanx = sinx / cosx
    tanx' = cox / -sinx <-------Is that right?

    If yes, then how am I suppose to make my equation by using "cox / -sinx "?
    By plugging that in... I get this:

    (((1 * tanX) - ((cox / -sinx) * X))) / ((tanx^2))
    Then I'm really stuck on that one... I mean, if I do plug in the "pi/3" to the "x" variables then it will just be a mess. Unless that's the only way to get the answer? Or should have I used the product rule instead?


    3)

    If f(x) = sinxcosx, then f '(pi/6)
    Product rule:
    y= f x g
    F'g + g'f
    Therefore...
    = (cosx*cosx) + (-sinx*sinx)
    = (cos^2x) + (-sin^2x)
    = (cos^2(pi/6)) + (-sin^2(pi/6))
    = 0.633

    -Let * be a multiplication sign
    -Is that right?

    4)
    Differentiate: f(x) = x^2 + 2tanx

    So, I'll just get the derivative of the equation...
    2x + 2(???)

    -Let ??? be "I don't know".
    So, I'm stuck. I have no idea what's the derivative of tanx have. I already encountered this problem in question #3 and I assumed that it would be:

    tanx = sinx / cosx
    tanx' = cox / -sinx

    Is that right? If yes, then I would get this equation:
    2x + 2(cosx/-sinx)

    Is that right? If yes, can I simplify it much more?

    5)
    Find the equation of the tangent line to the graph of f(x) = (x-1) / (x+1) when x = 1
    y=mx+b
    Chain Rule:
    = 1(x+1) - 1(x-1) / (x+1)^2
    = x+1 -x +1 / (x+1)^2
    = 2 / (x+1)^2
    = Plug in "x"
    = 2 / (1+1)^2
    = 2 / 4
    = 1/2
    slope (m) = 1/2

    Now that I have the slope, I'll just get the x & y values from plugging in "1" to the equation.
    y = (1-1) / (1+1)
    y = 0/2
    y = 0
    So: x = 1 and y = 0
    Then reflect on the slope equation:
    y = mx+b
    Plug in the numbers from what I have gotten before:
    0 = 1/2(1)+b
    0 - 1/2 = b
    -1/2 = b

    So...
    y = 1/2(x) + (-1/2)
    I'll multiply the whole equation by "2" to make it more neater.
    2y = 2(1/2x) + 2(-1/2)
    2y = x -1

    Is that right?



    ------------------------------------------
    I know my solutions were kind of long. But I hope that you could help me. I really want to make this happen. Or at least answer the questions correctly or in a much simplified way. =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2008
    Posts
    2
    Help, please.
    Please.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. Differentiation and partial differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 30th 2010, 10:16 PM
  3. Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 4th 2010, 10:45 AM
  4. Differentiation Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 25th 2010, 06:20 PM
  5. Differentiation and Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2009, 04:07 AM

Search Tags


/mathhelpforum @mathhelpforum