已知 Y=f((3x-2)÷(3x+2)) ,f·(x)=arctanx² 则 dy/dx x=0时 等于多少
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已知 Y=f((3x-2)÷(3x+2)) ,f·(x)=arctanx² 则 dy/dx x=0时 等于多少?
![已知 Y=f((3x-2)÷(3x+2)) ,f·(x)=arctanx² 则 dy/dx x=0时 等于多少](/uploads/image/z/2861095-31-5.jpg?t=%E5%B7%B2%E7%9F%A5+Y%3Df%EF%BC%88%EF%BC%883x-2%EF%BC%89%C3%B7%EF%BC%883x%2B2%EF%BC%89%EF%BC%89+%2Cf%C2%B7%EF%BC%88x%EF%BC%89%3Darctanx%26%23178%3B+%E5%88%99+dy%2Fdx+x%3D0%E6%97%B6+%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91)
由题意得:y=((3x-2)/(3x+2))=arctan((3x-2)/(3x+2))^2
则dy/dx=1/[1+(3x-2)/(3x+2))^2]×[(3(3x+2)-3(3x-2))/(3x+2)^2]
=(3(3x+2-3x+2))/[(3x+2)^2+(3x-2)^2]=12/[(3x+2)^2+(3x-2)^2]
则当x=0时,dy/dx=12/[2^2+(-2)^2]=12/8=3/2
再问: 答案是3/4 π
则dy/dx=1/[1+(3x-2)/(3x+2))^2]×[(3(3x+2)-3(3x-2))/(3x+2)^2]
=(3(3x+2-3x+2))/[(3x+2)^2+(3x-2)^2]=12/[(3x+2)^2+(3x-2)^2]
则当x=0时,dy/dx=12/[2^2+(-2)^2]=12/8=3/2
再问: 答案是3/4 π
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