作业帮判断y= x*sin(1x)可导性
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 12:16:51
y=Asin（ωx+ψ）周期为：T=2π/ωy=sin(x+1),ω=1,所以T=2π
y=cos²(x-π/12)^2+sin²(x+π/12)-1=(1/2)[1+cos(2x-π/6)+1-cos(2x+π/6)]-1=(1/2)[2sin2xsinπ/6]=(
记g(x)=f(x^2+sin^2x)+f(arctanx)=yg'(x)=f'(x^2+sin^2x)(2x+sin2x)+f'(arctanx)/(x2+1)dy/dx|x=0,即g'(0)代入得
y'=f'(sin²x)*(sin²x)'+f'(cos²x)*(cos²x)'=f'(sin²x)*(2sinxcos)+f'(cos²x
见图,复合函数求导.
dy/dx=2xf'(x²))cosf(x²)再问：没有过程吗？再答：复合函数求导法则
d/dx(f(sin^2(x))+sin(f(x)^2)) = sin(2 x) f'(sin^2(x))+2 f(x) f'
奇函数.f(x)＝lg[sinx+√(1+sin^2x)]因为[－sinx+√(1+sin^2x)]×[sinx+√(1+sin^2x)]＝1,所以,－sinx+√(1+sin^2x)＝1/[sinx
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
symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)
dx/dt={1/[2√(1-t^2)]}(-2t)=-t/√(1-t^2)dy/dt=1/√(1-t^2)dy/dx=[1/√(1-t^2)]/[-t/√(1-t^2)]=-1/t再问：为啥dy/d
先画y＝sinx.y＝－sinx关于x轴对称,再画y＝－sinx＋1,将图像向上平移一个单位就可以了.
答：y=sin(2x+3兀/2)y=sin(2x+2兀-1/2*兀)y=sin(2x-兀/2)y=-sin(兀/2-2x)y=-cos(2x)y=-(cosx)^2+(sinx)^2所以f(-x)=-
对这样的隐函数求导数的时候,就把y看作x的函数,y对x求导就得到dy/dx所以原等式对x求导得到2xy²+x²*2y*dy/dx+siny+x*cosy*dy/dx=0于是化简得到
若看不清楚,可点击放大.
按定义做.lim(x-->0)|(f(x)-f(0))/x-0|=lim(x-->0)|x^2sin(1/x)|/|x|=lim(x-->0)|xsin(1/x)|因为|sin(1/x)|0)|xsi
这个是复合函数的求导问题dy/dx=f'(sin^2x)*(sin^2x)'+f'(cos^2x)*(cos^2x)'=f'(sin^2x)(2sinx*cosx)+f'(cos^2x)*(-2cos
y=3^[sin(1/x)]y'=3^[sin(1/x)]ln3*cos(1/x)*(-1/x^2)=-ln3*3^[sin(1/x)]*cos(1/x)/x^2
y'=f'(sin(2x))*(sin(2x))'+(sin(f(2x)))'*f'(2x)=f'(sin(2x))*2*cos(2x)+cos(f(2
f(x)={sin(-x)/x,x