Wie findet man f '(x) anhand der Definition einer Ableitung f (x) = sqrt (x - 3)?

Wie findet man f '(x) anhand der Definition einer Ableitung f (x) = sqrt (x - 3)?
Anonim

Antworten:

Nutzen Sie einfach die Vorteile # a ^ 2-b ^ 2 = (a-b) (a + b) #

Antwort ist:

#f '(x) = 1 / (2sqrt (x-3)) #

Erläuterung:

#f (x) = sqrt (x-3) #

#f '(x) = lim_ (h-> 0) (sqrt (x + h-3) -sqrt (x-3)) / h = #

# = lim_ (h -> 0) ((sqrt (x + h-3) -sqrt (x-3)) * (sqrt (x + h-3) + sqrt (x-3))) / (h (sqrt (x + h-3) + sqrt (x-3))) = #

# = lim_ (h -> 0) (sqrt (x + h-3) ^ 2-sqrt (x-3) ^ 2) / (h (sqrt (x + h-3) + sqrt (x-3))) = #

# = lim_ (h -> 0) (x + h-3-x-3) / (h (sqrt (x + h-3) + sqrt (x-3))) = #

# = lim_ (h-> 0) h / (h (sqrt (x + h-3) + sqrt (x-3))) = #

# = lim_ (h-> 0) cancel (h) / (cancel (h) (sqrt (x + h-3) + sqrt (x-3))) = #

# = lim_ (h-> 0) 1 / ((sqrt (x + h-3) + sqrt (x-3))) = #

# = 1 / ((sqrt (x + 0-3) + sqrt (x-3))) = 1 / (sqrt (x-3) + sqrt (x-3)) = #

# = 1 / (2sqrt (x-3)) #