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]#