# What is the derivative of e^x from first principles.

### 1 Answer | Add Yours

`(d[e^x])/(dx) = lim_(h-gt0) (e^(x+h)-e^x)/h`

`(d[e^x])/(dx) = lim_(h-gt0) e^x[(e^h-1)/h]`

`(d[e^x])/(dx) = e^xlim_(h-gt0)[(e^h-1)/h]`

Let, `h = ln(1+1/n)`

Then, when `h-gt0 , n -gtoo`

`(d[e^x])/(dx) = e^xlim_(n-gtoo)[(e^ln(1+1/n)-1)/(ln(1+1/n))]`

`(d[e^x])/(dx) = e^xlim_(n-gtoo)[(1/n)/(ln(1+1/n))]`

`(d[e^x])/(dx) = e^x[1/(lim_(n->oo)nxxln(1+1/n))]`

`(d[e^x])/(dx) = e^x[1/(lim_(n->oo)ln(1+1/n)^n)]`

`(d[e^x])/(dx) = e^x(1/(ln[lim_(n->oo)(1+1/n)^n]))`

By definition of e, `lim_(n->oo)(1+1/n)^n = e`

`(d[e^x])/(dx) = e^x(1/(ln[e]))`

Therefore,

`(d[e^x])/(dx) = e^x`

**Sources:**