What is the second derivative of $f(x)=cos(x^2)$ ?

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Answer 1 · 2018-05-15T16:36:21

$(d^2f)/(dx^2)=-2sin(x^2)-2x^2cos(x^2)$

We use chain rule here

As $f(x)=cos(x^2)$

$(df)/(dx)=-sin(x^2)xx d/(dx)x^2=-sin(x^2)xx2x=-2xsin(x^2)$

and $(d^2f)/(dx^2)=-1*2sin(x^2)-x[cos(x^2)*2x]$

= $-2sin(x^2)-2x^2cos(x^2)$

Answer 2 · 2018-05-15T16:48:25

Here,

$f(x)=cosx^2$

Diff.w.r.t., $x$ , $"using "color(blue)"Chain Rule:"$

$f'(x)=-sinx^2 (d/(dx)(x^2))$

$=>f'(x)=-sinx^2(2x)$

$=>f'(x)=-2[xsinx^2]$

Again diff.w.r.t. $x$ , $" using "color(blue)"product Rule :"$ ,

$f''(x)=-2[xcosx^2*2x+sinx^2]$

$f''(x)=-2[2x^2cosx^2+sinx^2]$

What is the second derivative of f(x)=cos(x^2) f(x)=cos(x^2) ?