高数,导数是对X求导还是对fai(x)求导
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高数,导数是对X求导还是对fai(x)求导
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![高数,导数是对X求导还是对fai(x)求导](/uploads/image/z/16523520-24-0.jpg?t=%E9%AB%98%E6%95%B0%2C%E5%AF%BC%E6%95%B0%E6%98%AF%E5%AF%B9X%E6%B1%82%E5%AF%BC%E8%BF%98%E6%98%AF%E5%AF%B9fai%EF%BC%88x%EF%BC%89%E6%B1%82%E5%AF%BC)
12.设函数y=y(x)由方程x²y+y²lnx-1=0所确定,求y'(1).
解一:方程两边对x求导:2xy+x²y'+2yy'lnx+(y²/x)=0,(x²+2ylnx)y'=-2xy-(y²/x)=-(2x²y+y²)/x
故y'=-(2x²y+y²)/(x³+2xylnx);
解二:设F(x,y)=x²y+y²lnx-1=0,则
y'=dy/dx=-(∂F/∂x)/(∂F/∂y)=-(2xy+y²/x)/(x²+2ylnx)=-(2x²y+y²)/(x³+2xylnx);
x=1时y=1;将坐标(1,1)代入得y'(1)=-2.
13.设y=y(x)由方程lny=(x+y)/x所确定,求y'.
解一.两边对x求导:y'/y=[x(1+y')-(x+y)]/x²,由此式解出y'=y²/[x(y-x)];
解二.设F(x,y)=[(x+y)/x]-lny=0
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-{[x-(x+y)]/x²}/[(1/x)-(1/y)]=(y/x²)/[(y-x)/xy]=y²/[x(y-x)];
14.设方程y^x=(y+x)确定了函数y=y(x),求dy/dx.
两边取自然对数得xlny=ln(y+x);设F(x,y)=xlny-ln(y+x)=0,
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-[(lny)-1/(y+x)]/[(x/y)-1/(y+x)]=y[1-(y+x)lny]/(xy+x²-y)
15.设φ(x)>0处处可微,求d{φ[ln(φ(x))/φ(x)]}
d{φ[ln(φ(x))/φ(x)]}={φ'[ln(φ(x))/φ(x)]}{[φ'(x)-φ'(x)ln(φ(x))]/φ²(x)}dx
16.设函数y=y(x)由tany=(e^x)+y所确定,求dy.
两边对x求导得(sec²y)y'=(e^x)+y';(sec²y-1)y'=e^x,故y'=(e^x)/(sec²y-1)
∴dy=[(e^x)/(sec²y-1)]dx
17.设y=y(x)由方程e^(x-y)-xsiny=1确定,求dy
两边对x求导得:[e^(x-y)](1-y')-siny-xy'cosy=0
[e^(x-y)+xcosy]y'=e^(x-y)-siny,故y'=[e^(x-y)-siny]/[e^(x-y)+xcosy];
∴dy= {[e^(x-y)-siny]/[e^(x-y)+xcosy]}dx
18.题目不全,没法帮你.
再问: 15题为什么不是对X求导呢
再答: 当然是对x求导啦!那是一个复合函数:设φ(x)>0处处可微,求d{φ[ln(φ(x))/φ(x)]}; 可以这样分设y=φ(u),u=(lnv)/v,v=φ(x);那么dy/dx=(dy/du)(du/dv)(dv/dx).
解一:方程两边对x求导:2xy+x²y'+2yy'lnx+(y²/x)=0,(x²+2ylnx)y'=-2xy-(y²/x)=-(2x²y+y²)/x
故y'=-(2x²y+y²)/(x³+2xylnx);
解二:设F(x,y)=x²y+y²lnx-1=0,则
y'=dy/dx=-(∂F/∂x)/(∂F/∂y)=-(2xy+y²/x)/(x²+2ylnx)=-(2x²y+y²)/(x³+2xylnx);
x=1时y=1;将坐标(1,1)代入得y'(1)=-2.
13.设y=y(x)由方程lny=(x+y)/x所确定,求y'.
解一.两边对x求导:y'/y=[x(1+y')-(x+y)]/x²,由此式解出y'=y²/[x(y-x)];
解二.设F(x,y)=[(x+y)/x]-lny=0
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-{[x-(x+y)]/x²}/[(1/x)-(1/y)]=(y/x²)/[(y-x)/xy]=y²/[x(y-x)];
14.设方程y^x=(y+x)确定了函数y=y(x),求dy/dx.
两边取自然对数得xlny=ln(y+x);设F(x,y)=xlny-ln(y+x)=0,
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-[(lny)-1/(y+x)]/[(x/y)-1/(y+x)]=y[1-(y+x)lny]/(xy+x²-y)
15.设φ(x)>0处处可微,求d{φ[ln(φ(x))/φ(x)]}
d{φ[ln(φ(x))/φ(x)]}={φ'[ln(φ(x))/φ(x)]}{[φ'(x)-φ'(x)ln(φ(x))]/φ²(x)}dx
16.设函数y=y(x)由tany=(e^x)+y所确定,求dy.
两边对x求导得(sec²y)y'=(e^x)+y';(sec²y-1)y'=e^x,故y'=(e^x)/(sec²y-1)
∴dy=[(e^x)/(sec²y-1)]dx
17.设y=y(x)由方程e^(x-y)-xsiny=1确定,求dy
两边对x求导得:[e^(x-y)](1-y')-siny-xy'cosy=0
[e^(x-y)+xcosy]y'=e^(x-y)-siny,故y'=[e^(x-y)-siny]/[e^(x-y)+xcosy];
∴dy= {[e^(x-y)-siny]/[e^(x-y)+xcosy]}dx
18.题目不全,没法帮你.
再问: 15题为什么不是对X求导呢
再答: 当然是对x求导啦!那是一个复合函数:设φ(x)>0处处可微,求d{φ[ln(φ(x))/φ(x)]}; 可以这样分设y=φ(u),u=(lnv)/v,v=φ(x);那么dy/dx=(dy/du)(du/dv)(dv/dx).
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