Rinkoo Gupta
Last Activity: 11 Years ago
f(x)=(cosx+5cos3x+cos5x)/cos6x+6cos4x+15cos2x+10)
f(0)=(1+5+1)/(1+6+15+10)=7/32
f'(x)={(cos6x+6cos4x+15cos2x+10)(-sinx-15sin3x-5sin5x)-(cosx+5cos3x+cos5x)(-6sin6x-24sin4x-30sin2x)}/(cos6x+6cos4x+15cos2x+10)^2
=>f(0)={(32)(0)-(7)(0)}/(32)^2=0
f"(x)=[(cos6x+6cos4x+15cos2x+10)^2 {(cos6x+6cos4x+15cos2x+10)(-cosx-45cos3x-25cos5x)+(-sinx-15sin3x-5sin5x)(-6sin6x-24sin4x-30sin2x)+(cosx+5cos3x+cos5x)(36cos6x+96cos4x+60cos2x)}-{cos6x+6cos4x+15cos2x+10)(-sinx-15sin3x-5sin5x)+(cosx+5cos3x+cos5x)(6sin6x+24sin4x+30sin2x).2(cos6x+6cos4x+15cos2x+10)(-6sin6x-24sin4x-30sin2x)]/(c0s6x+6cos4x+15cos2x+10)^4
f"(0)=[(32)^2{(32)(-71)+(7)(152)}-{(32)(0)+0}]/(32)^4
=(32)^2{-2272+1064}/(32)^4
=-1208/1024
f(0)+f'(0)+f"(0)=7/32+0-1208/1024
=7/32-1208/1024
=(224-1208)/1024
=-984/1024
Thanks & Regards
Rinkoo Gupta
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