y=(1-x^2)/(1+x^2)的值域导数法!)答案为(-1,1]令f(x)=1-x^2,g(x)=1+x^2,有f'(x)=-2x,g'(x)=2xy=f(x)/g(x)=(f'g-fg')/g^2=(-2x*(1+x^2)-(1-x^2)*2x)/(1+x^2)^2=(-4x)/(1+x^2)^2
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![y=(1-x^2)/(1+x^2)的值域导数法!)答案为(-1,1]令f(x)=1-x^2,g(x)=1+x^2,有f'(x)=-2x,g'(x)=2xy=f(x)/g(x)=(f'g-fg')/g^2=(-2x*(1+x^2)-(1-x^2)*2x)/(1+x^2)^2=(-4x)/(1+x^2)^2](/uploads/image/z/1042553-65-3.jpg?t=y%3D%281-x%5E2%29%2F%281%2Bx%5E2%29%E7%9A%84%E5%80%BC%E5%9F%9F%E5%AF%BC%E6%95%B0%E6%B3%95%21%29%E7%AD%94%E6%A1%88%E4%B8%BA%28-1%2C1%5D%E4%BB%A4f%28x%29%3D1-x%5E2%2Cg%28x%29%3D1%2Bx%5E2%EF%BC%8C%E6%9C%89f%27%28x%29%3D-2x%2Cg%27%28x%29%3D2xy%3Df%28x%29%2Fg%28x%29%3D%28f%27g-fg%27%29%2Fg%5E2%3D%28-2x%2A%281%2Bx%5E2%29-%281-x%5E2%29%2A2x%29%2F%281%2Bx%5E2%29%5E2%3D%28-4x%29%2F%281%2Bx%5E2%29%5E2)
y=(1-x^2)/(1+x^2)的值域导数法!)答案为(-1,1]令f(x)=1-x^2,g(x)=1+x^2,有f'(x)=-2x,g'(x)=2xy=f(x)/g(x)=(f'g-fg')/g^2=(-2x*(1+x^2)-(1-x^2)*2x)/(1+x^2)^2=(-4x)/(1+x^2)^2
y=(1-x^2)/(1+x^2)的值域导数法!)答案为(-1,1]
令f(x)=1-x^2,g(x)=1+x^2,有f'(x)=-2x,g'(x)=2x
y=f(x)/g(x)=(f'g-fg')/g^2=(-2x*(1+x^2)-(1-x^2)*2x)/(1+x^2)^2=(-4x)/(1+x^2)^2
y=(1-x^2)/(1+x^2)的值域导数法!)答案为(-1,1]令f(x)=1-x^2,g(x)=1+x^2,有f'(x)=-2x,g'(x)=2xy=f(x)/g(x)=(f'g-fg')/g^2=(-2x*(1+x^2)-(1-x^2)*2x)/(1+x^2)^2=(-4x)/(1+x^2)^2
这题求最大值可以用导数,根据你求的导数
(1+x^2)^2恒大于0
所以当想x
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