5. Hence show that: OU (hvexp(-ßhv)) hv exp(-2ẞhv) =-3Nhv + οβ (1-exp(-ẞhv)) (1-exp(-Bhv)) Hint 1: you will need to use the quotient rule: f'(x) yf'(x) xf'(y) f'(y) (f'(y))² (f'(y))²
5. Hence show that: OU (hvexp(-ßhv)) hv exp(-2ẞhv) =-3Nhv + οβ (1-exp(-ẞhv)) (1-exp(-Bhv)) Hint 1: you will need to use the quotient rule: f'(x) yf'(x) xf'(y) f'(y) (f'(y))² (f'(y))²
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![5. Hence show that:
OU
(hv exp(-Bhv))
hv exp(-2ẞhv)
=-3Nhv
+
οβ
(1-exp(-ẞhv)) (1-exp(-ßhv))²
Hint 1: you will need to use the quotient rule:
yf'(x) xf'(y)
f'(x)
f'(y) (f'(y))² (f'(y))²
Hint 2: the following identity will avoid you having to use the chain rule:
d
(1-e√(x)) = -f'(x)e√(x)
dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4dc86bbc-2ec8-4b12-bfec-32681d151b9f%2F8a8571fb-8d24-410b-9bef-d6bd8623a3da%2Flc7rau_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. Hence show that:
OU
(hv exp(-Bhv))
hv exp(-2ẞhv)
=-3Nhv
+
οβ
(1-exp(-ẞhv)) (1-exp(-ßhv))²
Hint 1: you will need to use the quotient rule:
yf'(x) xf'(y)
f'(x)
f'(y) (f'(y))² (f'(y))²
Hint 2: the following identity will avoid you having to use the chain rule:
d
(1-e√(x)) = -f'(x)e√(x)
dx
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