Question

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

Answer 1

`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!) + ...`