How can I find the derivative of #y=e^x# from first principles?
1 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
And so:
# d/dx e^x=e^x#
This special exponential function with Euler's number,
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!) + ... #