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F Cosx Cos2x

f(cosx) = cos2x = 2(cosx)^2-1 => f(x) = 2x^2 -1 f(sin15°)= 2(sin15°)^2 -1 = -(1-2(sin15°)^2) = - cos30° =-√3/2

解:设t=2-cosx∈[1,3]cosx=2-tcos2x=2(cosx)^2-1=2(t-2)^2-1=2t^2-8t+7故有:f(t)=2t^2-8t+7+2-t=2t^2-9t+9,t∈[1,3]f(x-1)=2(x-1)^2-9(x-1)+9=2x^2-4x+4-9x+9+9=2x^2-13x+22(x∈[2,4])

f'(cosx)=cos2x=2cos^2x-1 故f'(x)=2x^2-1 即f''(x)=(2x^2-1)'=4x。

不详

f(x)=cosx+cos2x =cosx+2cosx-1 =3cosx-1 因为-1

f'(cosx)=cos2x=2(cosx)^2-1 f'(x)=2x^2-1 f''(x)=4x

题干不全,无法作答

先用2^n·sinx乘表达式的两边,求得f(x)=sin2nx/(2^n·sinx),然后再根据商的求导法则和复合函数求导法则,很容易求出f''(x),进而求出f''(0).

f(cosX)=cos2X f(cosx)=2cos²x-1,令cosx=X ,则原来的函数解析式为:f(X)=2X²-1,所以f(sin5派/12)=2[sin(5π/12﹚]²-1又∵sin(5π/12)=(√6 +√2)/4,∴f(sin5π/12)=(√3 -2)/4.

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