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设z=(1+x^2y)^y,求&z/&x,&z/&y
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设z=(1+x^2y)^y,求&z/&x,&z/&y
▼优质解答
答案和解析
lnz=yln(1+yx^2)
∂lnz/∂x=(1/z)∂z/∂x=2xy^2/(1+yx^2)
∂z/∂x=2zxy^2/(1+yx^2)=2xy^2(1+yx^2)^(y-1)
∂lnz/∂y=(1/z)∂z/∂y=ln(1+yx^2)+yx^2/(1+yx^2)
∂z/∂y=(1+yx^2)^y [ln(1+yx^2) + yx^2/(1+yx^2)]
∂lnz/∂x=(1/z)∂z/∂x=2xy^2/(1+yx^2)
∂z/∂x=2zxy^2/(1+yx^2)=2xy^2(1+yx^2)^(y-1)
∂lnz/∂y=(1/z)∂z/∂y=ln(1+yx^2)+yx^2/(1+yx^2)
∂z/∂y=(1+yx^2)^y [ln(1+yx^2) + yx^2/(1+yx^2)]
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