From First Principles

Key Questions

  • Answer:

    # d/dx e^x = e^x #

    Explanation:

    We seek:

    # d/dx e^x#

    Method 1 - Using the limit definition:

    # f'(x) = lim_{h to 0} {f(x+h)-f(x)}/{h} #

    We have:

    # f'(x) = lim_{h to 0} {e^(x+h)-e^(x)}/{h} #
    # " " = lim_{h to 0} {e^xe^h-e^(x)}/{h} #
    # " " = lim_{h to 0} e^x((e^h-1))/{h} #
    # " " = e^xlim_{h to 0} ((e^h-1))/{h} #

    Think about this limit for a moment and we can rewrite it as:

    #lim_{h to 0} ((e^h-1))/{h} = lim_{h to 0} ((e^h-e^0))/{h} #
    # " " = lim_{h to 0} ((e^(0+h)-e^0))/{h} #
    # " " = f'(0) # (by the derivative definition)

    Hence,

    # f'(x) = e^xf'(0) #

    Now, It can be shown that this limit:

    # f'(0) = lim_{h to 0} ((e^h-1))/{h} #

    both exists and is equal to unity. Additionly, the number #2.718281 ...#, which we call Euler's number) denoted by #e# is extremely important in mathematics, and is in fact an irrational number (like #pi# and #sqrt(2)#,

    And so:

    # d/dx e^x=e^x#

    This special exponential function with Euler's number, #e#, is the only function that remains unchanged when differentiated.

    Method 2 - Power Series

    We can use the power series:

    # e^x = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #

    Then we can differentiate term by term using the power rule:

    # d/dx e^x = d/dx{1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... } #

    # \ \ \ \ \ \ \ \ \ = 0 +1 + (2x)/(2!) + (3x^2)/(3!) + (4x^3)/(4!) + (5x^4)/(5!) + ... #

    # \ \ \ \ \ \ \ \ \ = 1 + (x)/(1!) + (3x^2)/(2! * 2) + (4x^3)/(3! * 4) + (5x^4)/(4! * 5) + ... #

    # \ \ \ \ \ \ \ \ \ = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #

  • By rewriting a bit,

    #y=c^x=e^{(lnc)x}#.

    By Chain Rule,

    #y'=e^{(lnc)x}cdot(lnc)=(lnc)c^x#


    I hope that this was helpful.

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