¿Cómo se diferencian sqrt ((x + 1) / (2x-1))?

Responder:

# - (3 (x + 1)) / (2 (2x-1) ^ 2 sqrt ((x + 1) / (2x-1)) #

Explicación:

#f (x) = u ^ n #

#f '(x) = n xx (du) / dx xxu ^ (n-1) #

En este caso:# sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ (1/2): #

#n = 1/2, u = (x + 1) / (2x-1) #

# d / dx = 1/2 xx (1xx (2x-1) - 2xx (x + 1)) / (2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 / 2-1) #

# = 1 / 2xx (-3) / ((2x-1) ^ 2 xx ((x + 1) / (2x-1)) ^ (1 / 2-1) #

# = - (3 (x + 1)) / (2 (2x-1) ^ 2 ((x + 1) / (2x-1)) ^ (1/2) #

Qué es (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt (3-) sqrt (5))?

2/7 Tomamos, A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) -sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5-sqrt3) / (2sqrt3-loes-lo-las-condiciones de la palabra-sqrt5-sqrt3) ) (2sqrt3 + sqrt5)) / ((2sqrt3 + sqrt5) (2sqrt3-sqrt5) = ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt15) ^ 2- (sqrt5) ^ 2) = (cancel (2sqrt15) -5 + 2 * 3cancel (-sqrt15) - cancel (2sqrt15) -5 + 2 * 3 + cancel (sqrt15)) / (12-5) = ( -10 + 12) / 7 = 2/7 Tenga en cuenta que si en los denominadores son (sqrt3 + sqrt (3 + sqrt5)) y (sq

¿Cómo se diferencian sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?

(dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy ) / (dx) = 1 / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) * sen (x ^ 2 + 2) * 2x + 2sen (x + 2) (dy ) / (dx) = (2xsen (x ^ 2 + 2) + 2sen (x + 2)) / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (cancel2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (cancel2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)))

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

F '(x) == - (sqrt (e ^ cot (x)). csc ^ 2 (x)) / 2 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)" Sea u (x) = cot (x) => u' (x) = -csc ^ 2 (x) yg (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 ^ cuna (x )) e ^ cot (x) .- cos ^ 2 (x) = (- e ^ cot (x) csc ^ 2