Question

Find the derivative of the following function from the first principle.

e^x

Answer 1

Let f(x) = e^-x then f(x + h) = `e^(-(x+h))`

Now `"d"/"dx"` f(x)

`= lim_(h->0) ("f"(x + "h") - "f"(x))/"h"`

`= 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) ((1/e^"h" - 1)/"h")`

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

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

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

`= -e^(-x) 1/e^0 xx 1`

`[because lim_(x->0) (e^x - 1)/x = 1]`

`"d"/"dx"` f(x) = `- e^-x`

`therefore "d"/"dx" (e^-x) = - e^-x`