¿Cómo se diferencian f (x) = sqrt (e ^ cot (x)) usando la regla de la cadena?

Responder:

f '(x) == -# (sqrt (e ^ cot (x)). csc ^ 2 (x)) / 2 #

Explicación:

#f (x) = sqrt (e ^ cot (x)) #

Para encontrar la derivada de f (x), necesitamos usar la regla de la cadena.

#color (rojo) "regla de la cadena: f (g (x)) '= f' (g (x)). g '(x)" #

Dejar #u (x) = cot (x) => u '(x) = - csc ^ 2 (x) #

y # g (x) = e ^ (x) => g '(x) = e ^ (x).g' (u (x)) = e ^ cuna (x) #

#f (x) = sqrt (x) => f '(x) = 1 / (2sqrt (x)) => f' (g (u (x))) = 1 / (2sqrt (e ^ cot (x)) #

# d / dx (f (g (u (x))) = f '(g (u (x))). g' (u (x)). u '(x) #

=# 1 / (sqrt (e ^ cot (x))) e ^ cot (x).- cos ^ 2 (x) #

=# (- e ^ cot (x) csc ^ 2x) / sqrt (e ^ cot (x)) #

#color (azul) "cancela e ^ cot (x) con sqrt (e ^ cot (x)) en el denominador" #

=-# (sqrt (e ^ cot (x)). csc ^ 2 (x)) / 2 #

¿Cómo se distingue f (x) = sqrt (cote ^ (4x) usando la regla de la cadena?

F '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cuna (e ^ (4x))) ^ (- 1/2)) / 2 color (blanco) (f' (x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x)) f (x) = sqrt (cot (e ^ (4x))) color (blanco) (f (x)) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' (x) color (blanco ) (f '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 g (x) = cuna (e ^ (4x)) color (blanco) (g (x)) = cuna (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ (4x) color (blanco) (h ( x)) = e ^ (j (x)) h '(x) = j' (x) e ^ (j (x)) j (x) = 4x j '(x) = 4 h' (x) = 4e ^ (4x) g '(x) = - 4e ^ (4x) csc ^ 2 (e ^

¿Cómo se diferencia f (x) = sqrt (ln (x ^ 2 + 3) usando la regla de la cadena?

F '(x) = (x (ln (x ^ 2 + 3)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3))) Se nos da: y = (ln (x ^ 2 + 3) ) ^ (1/2) y '= 1/2 * (ln (x ^ 2 + 3)) ^ (1 / 2-1) * d / dx [ln (x ^ 2 + 3)] y' = ( ln (x ^ 2 + 3)) ^ (- 1/2) / 2 * d / dx [ln (x ^ 2 + 3)] d / dx [ln (x ^ 2 + 3)] = (d / dx [x ^ 2 + 3]) / (x ^ 2 + 3) d / dx [x ^ 2 + 3] = 2x y '= (ln (x ^ 2 + 3)) ^ (- 1/2) / 2 * (2x) / (x ^ 2 + 3) = (x (ln (x ^ 2 + 3)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3)))

¿Cómo se distingue f (x) = sqrt (ln (1 / sqrt (xe ^ x)) usando la regla de la cadena?

Solo encadena la regla una y otra vez. f '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) f (x) = sqrt (ln (1 / sqrt (xe ^ x))) Bien, esto va a ser difícil: f '(x) = (sqrt (ln (1 / sqrt (xe ^ x))))' = = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * (ln (1 / sqrt (xe ^ x)))) = = 1 / (2sqrt (ln (1 / sqrt (xe ^ x))))) * 1 / (1 / sqrt (xe ^ x)) (1 / sqrt (xe ^ x)) '= = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * sqrt (xe ^ x) (1 / sqrt (xe ^ x)) '= = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x))))) (1 / sqrt (xe ^ x))' = = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) ((xe ^