f(x)=2cosx10x−π22cosx1012cosx⇒f'(x)=
0
2
π2
π – 6
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Given : f(x)=2cosx10x−π22cosx1012cosx
⇒ f(x) = 2cosx(4cos2x – 1) – 1(2xcosx – πcosx)
f(x) = 8cos3x – 2cosx – 2xcosx + πcosx
∴ f '(x) = – 8 × 3cos2x⋅sinx + 2sinx – 2cosx + 2xsinx + π(– sinx)
f '(x) = – 24cos2x⋅sinx + 2sinx + 2xsinx – 2cosx – πsinx
& f′π2=0+2+2π2−0−π.
f′π2=2+π−π
f′π2=2
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