Let `f(x)` be a differentiable function satisfying `f(x)=int_(0)^(x)e^((2tx-t^(2)))cos(x-t)dt`, then find the value of `f''(0)`.

Answer

`0`

Explanation

We have `f(x)=int_(0)^(x)e^(x^(2)-(x-t)^(2))cos(x-t)dt`
`:.f(x)=int_(0)^(x)e^(x^(2)-(x-(x-t))^(2))cos(x-(x-t))dt`
`implies f(x)=int_(0)^(x)e^(x^(2)-t^(2))costdt`
`impliesf(x)=e^(x^(2))int_(0)^(x)e^(-t^(2))cost dt`…………1
Differentiating w.r.t `x` we get
`f'(x)=2xe^(x^(2))int_(0)^(x)e^(-t^(2))cost dt +cosx`
Again differentiating w.r.t `x`, we get
`f''(x)=2(e^(x^(2))+2x^(2)e^(x^(2)))int_(0)^(x)e^(-t^(2))costdt+2xcosx-sinx`
`=2(1+2x^(2))f(x)+2xcosx-sinx`
`:'f''(0)=-2f(0)+0-0=0 ( :' f(0)=0 "from" 1)`