Kako najdete derivat inverzne trigonomske funkcije f (x) = arcsin (9x) + arccos (9x)?

Kako najdete derivat inverzne trigonomske funkcije f (x) = arcsin (9x) + arccos (9x)?
Anonim

Tukaj / kako jaz delam to:

- Pustila bom nekaj # "" theta = arcsin (9x) "" # in nekaj # "" alpha = arccos (9x) #

  • Torej dobim, # "" sintheta = 9x "" # in # "" cosalpha = 9x #

  • Razlikujem implicitno tako:

    # => (costheta) (d (theta)) / (dx) = 9 "" => (d (theta)) / (dx) = 9 / (costheta) = 9 / (sqrt (1-sin ^ 2theta)) = 9 / (sqrt (1- (9x) ^ 2) #

- Naslednje, ločujem # cosalpha = 9x #

# => (- sinalpha) * (d (alfa)) / (dx) = 9 "" => (d (alfa)) / (dx) = - 9 / (sin (alfa)) = - 9 / (sqrt) (1-cosalpha)) = - 9 / sqrt (1- (9x) ^ 2) #

  • Na splošno # "" f (x) = theta + alpha #

  • Torej, #f ^ ('') (x) = (d (theta)) / (dx) + (d (alfa)) / (dx) = 9 / sqrt (1- (9x) ^ 2) -9 / sqrt (1- (9x) ^ 2) = 0 #