The function `f: RvecR` satisfies `f(x^2)dotf^(x)=f^(prime)(x)dotf^(prime)(x^2)` for all real `xdot` Given that `f(1)=1` and `f^(1)=8` , then the value of `f^(prime)(1)+f^(1)` is `2` b. `4` c. `6` d. 8

Options

A. 2
B. 4
C. 6
D. 8

Answer

C

Explanation

We have `f(x^(2)).f''(x)=f'(x^(2))" (1)"`
Differentiating w.r.t. x,
`2xf'(x^(2))f''(x)+f(x^(2))f'''(x)=f''(x)f'(x^(2))+2xf''(x^(2))f'(x)`
Putting x = 1, we get
`2f'(1)f''(1)+f(1)f'''(1)=f''(1).f'(1)+2f''(1)f'(1)`
`therefore" "2f'(1)f''(1)+1xx8=3f''(1)f'(1)`
`therefore" "f'(1)f''1)=8" (ii)"`
Putting x = 1 in (i)
`f(1)f''(1)=f'(1)f'(1)`
`therefore" "f''(1)=(f'(1))^(2)`
So from (ii) `f'(1).(f'(1))^(2)=8`
`rArr" "f'(1)=2 and f''(1)=4`
`rArr" "f'(1)+f''(1)=6`