设F=sin(5y) sin(x-1)arctan(x y)对y求导-搜答案-灵鹊做题网

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

设y=sin[x/(1=x)]求dy

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/