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Calculus problems

  1. May 2, 2003 #1
    what is the derivative of the following functions with respect to x:
    a) y = 5e^-4x ans: -20e ^-4x
    b) y = 1/2e^x^2 ans: xe^x^2
    c)y = x^4 e^x ans: 4x^3 e^x + x^4 e^x
    d) y = e^-x (x) ans: - (x + 1)e^-x/(x^2)
    e) y = (1 + e^x)^1/2 ans: e^x/[2(1 + e^x)^1/2]
    f) y = x + e ^ (x)^1/2 ans: 1 + [e ^(x)^1/2 / 2(x)^1/2]

    3a) what is the equation of the tangent to the curve at the specified
    point?
    a) y = e^2x at (1, e^2) ans: y = 2e^2x - e^2
    b) y = e^x^2/(x) at (1, e) ans: y = x

    4)a) what is d^2/y/dx^2

    a) y = x^2 e^-x ans: (x^2 -4x +2)e^-x
    b) y = 4xe^x^2 ans: (16 x^3 + 24 x)e^x^2
    c) y = e^-x sin x ans: -2e^-x cos x

    5) how do you show that if y = e^x cos 2x then
    d^2y/dx - 2(dy/dx) + 5y = 0?

    =====
    How do you evaluate log (subcript 2) 1/32?
    b) log (g) 32 + log g(16) ?
    c) log (2) 3
    how do you solve for x? are there restrictions?
    a) log (x) 25 = 2/3
    b) log(7) [x + 7] + log(7) [x-7] = 0
     
  2. jcsd
  3. May 3, 2003 #2
    Shouldn't this be in the Homework Help forum ?
     
  4. May 3, 2003 #3

    FZ+

    User Avatar

    What are you stuck on? What can you do?
     
  5. May 3, 2003 #4
    Just a hint:

    A function such as f(x) = e^x, you simply multiply the derivative of the exponent with the actual function. For example:

    f(x) = e^(4x)

    exponent is 4x and the derivative is 4 so:

    f'(x) = 4 * e^(4x) = 4e^(4x)

    That's the basic tenet of derivatives involving e.

    If you need anymore help, just tell us where you are having trouble.
     
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