# From first principles find the derivative of e^x

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### 1 Answer

The derivative of a function f(x) from first principles is `f'(x) = lim_(h->0)(f(x+h)-f(x))/h` .

For the function `f(x) = e^x` ,

`f'(x) = lim_(h->0)(e^(x+h) - e^x)/h`

= `lim_(h->0)(e^x*e^h - e^x)/h`

= `lim_(h->0)(e^x*(e^h - 1))/h`

= `e^x*lim_(h->0)(e^h - 1)/h`

Use the series expansion of e^h, `e^h = 1 + h + h^2/(2!) + h^3/(3!) + .... + h^n/(n!) + ...`

= `e^x*lim_(h->0)(1 + h + h^2/(2!) + h^3/(3!) + .... + h^n/(n!) + ... - 1)/h`

= `e^x*lim_(h->0)(h + h^2/(2!) + h^3/(3!) + .... + h^n/(n!) + ...)/h`

= `e^x*lim_(h->0)(1 + h/(2!) + h^2/(3!) + .... + h^(n-1)/(n!) + ...)`

Substituting h = 0

= `e^x*1`

= `e^x`

**This proves from first principles that the derivative of `e^x` is `e^x` .**