设F=sin(5y) sin(x-1)arctan(x y)对y求导
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/19 17:20:58
你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏/8代入原式子,即sin(∏/4+φ)=1或者-1,再用(-π
y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x
见图,复合函数求导.
y=sinx²+sin²x∴y'=cos(x²)*(x²)'+2sinx*(sinx)'=2x*cos(x²)+2sinxcosx=2x*cos(x&
dy/dx=2xf'(x²))cosf(x²)再问:没有过程吗?再答:复合函数求导法则
2kπ-π/2≤2x+π/3≤2kπ+π/2得:kπ-5π/12≤x≤kπ+π/12增区间是:[kπ-5π/12,kπ+π/12],其中k∈Zx∈[-π/6,π/6],则:2x+π/3∈[0,2π/3
d/dx(f(sin^2(x))+sin(f(x)^2)) = sin(2 x) f'(sin^2(x))+2 f(x) f'
两边对x求导:y'e^y+(1+y')cos(x+y)=0,1)这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1)求导:y"e^y+(y')^2e^y+y"cos(x+y)-(1
设函数f(x,y)=sin(x+y),那么f(0,xy)=(sinxy)应该是sin0+sinsy=0+sinxy=sinxy再问:limsinxy\2x=()补充x→0,y→3另外一道题
dy/dx=cos{f[sinf(x)]}*{f[sinf(x)]}'=cos{f[sinf(x)]}*f‘[sinf(x)]*[sinf(x)]’=cos{f[sinf(x)]}*f‘[sinf(x
1)由三角函数和差化积公式:f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6
这道题你先看sinx必然大于等于零吧,sin((1-y)x)也必然大于等于零的吧?整个函数都是大于等于0的吧?那么你只要找到可以让函数取到零的x和y就可以得到最小值0那么试着凑一下,y=1,x=pai
f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知:y=-3
若看不清楚,可点击放大.
1)f(x)=sin(2x+φ)一条对称轴是X=π/8则kπ+π/2=2*π/8+φ===>φ=kπ+π/4因为-π
2x+φ=kπ+π/2,x=(kπ+π/2-φ)/2(kπ+π/2-φ)/2=π/8当k=0时,φ=π/4
1.由f(x)=sin(2x+φ)一条对称轴是直线x=π/2可得:在x=π/2时,函数取极值.则2*π/2+φ=kπ+π/2(k∈Z)φ=kπ-π/2又-π
1=x?
(1)由sin(2α+β)=3sinβ,得sin[(α+β)+α]=3sin[(α+β)-α],sin(α+β)cosα+cos(α+β)sinα=3sin(α+β)cosα-3cos(α+β)sin
f(x)=sin(2x+a)是R上的偶函数有f(x)=f(-x);sin(2x+a)=sin(-2x+a)=cos(π/2-(-2x+a))=cos(π/2+2x-a)余弦函数为R上的偶函数,a=π/