From First Principles

Key Questions

  • How can I find the derivative of y=e^x from first principles?

    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!) + ...

    Steve M · 2 · Apr 4 2018
  • How can I find the derivative of y=c^x using first principles, where c is an integer?

    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.

    Wataru · · Oct 17 2014

Questions

Differentiating Exponential Functions