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f(x)=cos³x+sin³x,f'(x)=3cos²x(-sinx)+3sin²x(cosx)=3sinxcosx(sinx-cosx),f'(x)=0, i.e. x=k+2, or x=k+4, kz, x (2k, 4+2k), sinx>0, cosx>0, sinx-cosx<0, so f'(x) <0, f(x) monotonically decreasing;
x ( 4+2k , 2+2k ), sinx>0, cosx>0, sinx-cosx>0, so f'(x) >0, f(x) monotonically increasing;
x ( 2+2k , 2k ), sinx>0, cosx<0, sinx-cosx>0, so f'(x) <0, f(x) monotonically decreasing;
x ( 2k ,5 4+2k ),sinx<0,cosx<0,sinx-cosx>0,so f'(x) >0, f(x) monotonically increasing;
x(5 4+2k ,3 2+2k ),sinx<0,cosx<0,sinx-cosx<0, so f'(x) <0, f(x) monotonically decreasing;
x(3 2+2k ,2 +2k ),sinx<0,cosx>0,sinx-cosx<0,so f'(x) >0, f(x) monotonically increasing;
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y=(sinx)^4+(cosx)^4
sinx)^2+(cosx)^2]^2-2(sinxcosx)^2
1-1/2*(sin2x)^2
1-1/4*2(sin2x)^2
1-1/4(1-cos4x)
3 source tremor 4+1 4*cos4x
When 4x [2k - 2k ],x [(2k-1) volt only 4,k 2], the monotonous hail hall defeat increases.
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y=(sinx)^4+(cosx)^4
sinx)^2+(cosx)^2]^2-2(sinxcosx)^2
1-1 Jean Oak 2*(sin2x) 2
1-1/4*2(sin2x)^2
1-1 Frank 4 (1-cos4x).
3/4+1/4*cos4x
When 4x [2k - 2k], x [(2k-1) 4, k 2], the single Ruchang tone increases.
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sinx 4-cosx 4=(sinx 2+cosx 2):auspicious*(lead sinx 2-cosx 2)=sinx 2-cosx 2=-cos2x
This function increments the Huai drain on (- 2+2k, 2k) (k z).
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Derivative, obtainable (cos(sinx)).'sin(sinx)cosxwhen x (2k, 2k), sinx>0, sin(sinx)>0 when beats nucleus x (-2k, 2k), sinx "zen only 0, sin(sinx)<0
When x (-2+2k, 2+2k), cosx>0, when x (2+2k, 3 2+2k), cosx<0 can be combined to get the result (it is too troublesome to type out - just understand the idea.
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y=x-x^3
y'=1-3x 2=-3(x+root3 3)(x-root3 3)when-root3 3 x root3 chainfield 3, y'0, monotonous increase;
When x-root number 3 canopy shouts 3, or x root number 3 3, y'0, monotonous minus bright hall.
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y=sinx+cosxy'=cosx-sinx= 2[ 2cosx- 2 2sinx]= 2cos(x+ 4) Find the monotonically increasing interval y below'>=0,2k - 2 x + 4 2k + 2 2k -3 4 x 2k + 4 monotonically increasing band cultivation interval [2k -3 4 ,2k + stupid only 4] The following is the monotonically decreasing interval.
The solution of atom6417 is really ingenious.
Now I will give you a more general solution. >>>More
i=∫(secx)^3dx=(1/2)×(secxtanx+ln|secx+tanx|)+c
The process is as follows: >>>More
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x^3-6x^2y+11xy^2-6y^3
x^3-6x^2y+9xy^2) +2xy^2-6y^3) >>>More
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