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A few Derivative Problems

  1. Oct 8, 2008 #1

    Here are several problems (differentiation) i have attempted but not completely sure if they are correct


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

    differentiate y = sec(ex)

    3. The attempt at a solution

    y = sec(ex)

    y = sec(u)

    (where u = ex)

    dy/dx = tan(u)sec(u)ex

    tan(ex)sec(ex)ex

    If this is correct so far, can it be further factorized?:

    ex(tan sec(ex))




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

    differentiate y = sec(log(x))


    3. The attempt at a solution

    y = sec(log(x))

    y = sec(u)

    (where u = log(x))

    dy/dx = tan(u)sec(u) 1/x

    = (tan sec(log(x))) / x



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

    find the second derivative of f(x) = sin(2x) + cos(3x)

    3. The attempt at a solution

    f(x) = sin(2x) + cos(3x)

    f ' (x) = 2cos(2x) - 3sin(3x)

    f '' (x) = cos(2x) - 4sin(2x) + sin(3x) - 9cos(3x)
     
  2. jcsd
  3. Oct 8, 2008 #2

    Mark44

    Staff: Mentor

    No. Your previous result is correct, but the simplification is not. The expression tan(ex) sec(ex) ex is the product of its three factors. Your mistake was changing the product of the tangent and secant factors into a composition of them.
    Same comment as in the first question. Also, is the log function here the natural log? If so, you should use ln instead of log, which normally means log base 10.
    Your first derivative is correct, the the second derivative is not. You are apparently using the product rule on 2cos(2x) and 3sin(3x). You can use the product rule here, but it's more powerful than what you need, and it's easier to use the constant multiple rule (i.e., d/dx (k * f(x)) = k * d/dx(f(x)) = k*f'(x) ). Where you made your mistake was forgetting that d/dx(2) = 0 and d/dx(3) = 0 in the coefficients of the two terms in f'(x).
    Mark
     
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