Y = 5 ^ sqrt (s)의 미분은 무엇입니까?

대답:

# dy / (ds) = (log (5) 5 ^ sqrt (s)) / (2sqrt (s)) #

설명:

체인 사용하기:

f '(x) = h'(x) g '(h (x)) #f (x) = g

와:

# g (u) = 5 ^ u => g '(u) = log (5) 5 ^ u #

#h (x) = sqrt (x) => 1 / (2sqrt (x)) #

이것을 함께 사용하면:

# dy / (ds) = (log (5) 5 ^ sqrt (s)) / (2sqrt (s)) #

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

2/7 우리는 A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5 (sqrt5 + sqrt3) / (sqrt5 + sqrt3) / (2sqrt3 + sqrt3) / (2sqrt3-sqrt3) / (2sqrt3-sqrt3) (2sqrt3 + sqrt5)) / (2sqrt3 + sqrt5) (2sqrt3 + sqrt5) = (2sqrt15-5 + 2 * 3- sqrt15) - (2sqrt15 + 5-2 * 3sqrt15) (2sqrt15) - (2sqrt15) -5 + 2 * 3 + 취소 (sqrt15)) / (12-5) = ( (sqrt3 + sqrt (3 + sqrt5)) 및 (sqrt3 + sqrt (3-sqrt5)) 인 경우, 해답은 변경 될 것이다.

F (x) = sqrt (1 + log_3 (x))의 미분은 무엇입니까?

(1 + logx3)) / {2sqrt (1 + log_3x)} = ((d / dx) (1 + logx / log3)) / { 2sqrt (1 + log_3x)} = (1 / (xln3)) / {2sqrt (1 + log_3x)} = 1 / (2xln3sqrt (1 + log_3)

당신은 어떻게 단순화합니까 (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) div sqrt (a + 1) / ( (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)

(1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) / (a-1) sqrt (a + 1) - (a + 1) sqrt (a + 1) 1) + sqrt (a + 1)) / (sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a- 1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) sqrt (a-1))) = color ( (a + 1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a + 1) (a + 1) -sqrt (a + 1))) = 컬러 (적색) (a + 1) - (sqrt (a + 1) cdot sqrt (a-1) (a + 1) = sqrt (a + 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + 1) (a + 1) -sqrt (a + 1))) xx ((a + 1) + sqrt (a + 1) (a + 1)) / 색 (적색) (((1 + 1))) (a-1)) / (sqrt (a-1))) xx ((sqrt (a + 1) cdot sqrt (a-1) (a + 1) cdot sqrt (a-1)) xx s