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Basic differentiation questions

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Hello guys. I had to do some test corrections for my AP Calculus AB class. I have completed all of them besides the four below. Can anyone tell me where I go wrong?

    1. Differentiate y = (1+cosx)/(1-cosx)

    dy/dx = [(1-cosx)(sinx)-(1+cosx)(sinx)]/(1-cosx)^2 (quotient rule)
    = [(sinx - sinx cosx) - (sinx + sinx cosx)]/(1-cosx)^2 (distribution)
    = (-sinx cosx - sinx cosx)/(1-cosx)^2 (simplification)
    = (-2sinx cosx)/(1-cosx)^2 (simplification)

    Answer choices:

    a. -1
    b. -2 cscx
    c. 2 cscx
    d. (-2sinx)/(1-cosx)^2

    Choice D is the closest to my answer, however my answer multiplies cosx with the -2sinx.

    2. Differentiate y = sin(x+y)
    I did this by implicit differentiation:

    y' = cos(x+y)[(x)+dy/dx] (doesn't seem correct...)

    Choices are:

    A. 0
    b. [cos(x+y)]/[1-cos(x+y)]
    c. cos(x+y)
    d. 1

    9. Differentiate: y = (secx)^2 + (tanx)^2

    y' = [(2secx)(secx tanx)] + [(2tanx)(secx)^2] (product rule)
    = 2 ((secx)^2)tanx + 2 tanx (secx)^2
    = 2 (sec x)^2 + 3 tanx

    Choices are:

    a. 0
    b. tan x + (secx)^4
    c. ((secx)^2)((secx)^2 + (tan x)^2)
    d. 4 (secx)^2 tanx

    (I skipped Pre-Calculus, which was essentially a trigonometry class, so I had a particularly difficult time with this one)
  2. jcsd
  3. Oct 22, 2008 #2
    d/dx cosx = - sinx NOT sinx

    Didn't check this one properly, but d/dx x = 1 NOT x

    Check step 3.

    Hope that helps! :smile:
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