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Question
Find the derivative of the following function from the first principle.
ex
Solution
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`
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