隐函数求导训练题
First
\[已知y=f(x)
\]
\[则y^{2}=f(x) \cdot f(x) \\
则y^{2}为包含f(x)的复合函数
\\ (y^{2})'=[f^{2}(x)]' \cdot f'(x)
\]
\[\\
\therefore (y^{2})'=2y \cdot y'
\]
Second
对数求导法:先取对数,再求导数
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\]
\[若y=x^{\frac{1}{x}},\quad 求y'(x)即\frac{dy}{dx}=?
\]
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\]
\[y=x^{\frac{1}{x}} \Rightarrow lny=\frac{1}{x}lnx
\]
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\]
将\(y\)视作\(x\)的函数,对两边求导:
\[\\
(lny)'=(\frac{lnx}{x})'
\\
\]
隐函数求导之术与复合函数求导之术通用.则根据链式法则: \(\frac{d}{dx}=\frac{d}{dy} \cdot \frac{dy}{dx}\)
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\]
\[\Rightarrow
\frac{d(lny)}{dx}\cdot \frac{dy}{dx}=\frac{d(\frac{lnx}{x})}{dx}\]
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\]
\[\frac{1}{y}\cdot \frac{dy}{dx}= \frac{lnx'x-x'lnx}{x^{2}}
\]
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\]
\[\because (x)'=(x^{1})'=1
\]
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\]
\[\therefore \frac{dy}{dx}=\frac{1-lnx}{x^{2}}\cdot y
\]
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\]
\[\frac{dy}{dx}=\frac{1-lnx}{x^{2}}\cdot x^{\frac{1}{x}}
\]
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\]
\[\frac{dy}{dx}=x^{-2}\cdot x^{\frac{1}{x}}\cdot (1-lnx)
\]
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\]
\[\frac{dy}{dx}= x^{\frac{1}{x}-2} (1-lnx)
\]
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\]
\[\therefore y'(x)=x^{\frac{1}{x}-2} (1-lnx)
\]
Second
\[x^{4}+y^{5}+y^{3}-1=0,\quad 求y'(x)即\frac{dy}{dx}
\]
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\]
\[y^{5}+y^{3}=1-x^{4}
\\\]
将\(y\)看成\(x\)的函数,同时对两侧求导: \((y^{5}+y^{3})'=(1-x^{4})'\)
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\]
根据链式法则得:
\[\frac{d(y^{5}+y^{3})}{dx}\cdot \frac{dy}{dx} =\frac{d(1-x^{4})}{dx}
\]
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\]
\[(5y^{4}+3y^{2})\cdot \frac{dy}{dx} =-4x^{3}
\]
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\]
\[\frac{dy}{dx}=\frac{-4x^{3}}{5y^{4}+3y^{2}}
\]
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\]
\[\therefore y'(x)=-\frac{4x^{3}}{5y^{4}+3y^{2}}
\]
Third
\[e^{xy}=sin(x+y), \quad \frac{dy}{dx}=?
\]
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\]
\[设xy=k,\quad x+y=u
\]
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\]
\[(e^{k})'=(sinu)'
\]
because they both are composite function, therefore:
\[(e^k)'\cdot (k)'=(sinu)'\cdot (u)'\]
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\]
\[(e^k)'(k)'\Rightarrow e^{k}[(x)(y)]'=e^{k}[x'y+y'x]=e^{k}(y+x\frac{dy}{dx})
\]
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\]
\[(sinu)'(u)'=cos(x+y)(1+\frac{dy}{dx})
\]
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\]
\[\Rightarrow e^{xy}(y+x\frac{dy}{dx})=cos(x+y)(1+\frac{dy}{dx})
\]
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\]
\[e^{xy}y+e^{xy}x\frac{dy}{dx}=cos(x+y)+\frac{dy}{dx}cos(x+y)
\]
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\]
\[e^{xy}x\frac{dy}{dx}-\frac{dy}{dx}cos(x+y)=
cos(x+y)-e^{xy}y\]
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\]
\[\frac{dy}{dx}[e^{xy}x-cos(x+y)]=cos(x+y)-e^{xy}y
\]
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\]
\[\frac{dy}{dx}=\frac{cos(x+y)-e^{xy}y}{e^{xy}x-cos(x+y)}
\]
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\]
\[\therefore y'(x)=\frac{cos(x+y)-e^{xy}y}{e^{xy}x-cos(x+y)}
\]
Fourth
\[隐函数求导: \quad x^{2}-y^{2}-4xy=0
\\
2边同时求导: \quad (x^{2})'-(y^{2})'-(4xy)'=0
\]
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\]
\[ (x^{2})'=2x \\ (y^{2})'= (y^{2})' \cdot y' =2yy'
\\
(4xy)'=(4x)'y+4x(y)'=4y+4xy' \]
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\]
\[2x-2yy'-4y-4xy'=0
\\
-2yy'-4xy'=-2x+4y
\\
y'(-2y-4x)=-2x+4y\]
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\]
\[y'=\frac{-2x+4y}{-2y-4x}
\]
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\]
\[y'=\frac{-2(x-2y)}{-2(y+2x)}
\]
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\]
\[y'=\frac{x-2y}{2x+y}
\]
