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Suppose f and g are functions having sec...

Suppose f and g are functions having second derivatives `f' and g'` every where, if `f(x).g(x)=1` for all `x and f'', g''` are never zero then `(f''(x))/(f'(x))-(g''(x))/(g'(x))` equals

A

`3((f'')/(g)-(g'')/(f))`

B

`3((f'')/(f)-(g'')/(g))`

C

`3((g'')/(g)-(f'')/(g))`

D

`3((f'')/(g)-(g'')/(f))`

Text Solution

Verified by Experts

The correct Answer is:
B
We have,
`f(x)g(x)=1`
Differentiating with repect to s, we get
`f'g+f'g=0" "…(i)`
Ditterentiating (i) w.r.t. x, we get
`f''g+2f'g'+fg''=0" "...(ii)`
Differentiating (ii) w.r.t. x, we get
`implies" "f'''g+g'''f+3f''g'+3g''f'=0`
`(f''')/(f')(f'g)+(g''')/(g')(fg')+(3f'')/(f)(fg')+(3g'')/(g)(gf')=0`
`implies" "((f''')/(f')+(3g'')/(g))(f'g)=-((g''')/(g')+(3f'')/(f))(fg')`
`implies" "-((f''')/(f')+(3g'')/(g))(fg')=-((g''')/(g')+(3f'')/(f))fg'" "["Using (1)"]`
`implies" "(f''')/(f')+(3g'')/(g)=(g''')/(g')+(3f'')/(f)`
`implies" "(f''')/(f')=(g''')/(g')=3((f'')/(f)-(g'')/(g))`
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