求d^y\dx^
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求d^y\dx^
2、(1)y=sin[f(x²)],y'=cos[f(x²)]*f'(x²)*2x;
y"=-sin[f(x²)]f(x²)*2x+cos[f(x²)]*[f'(x²)*2x]'
=-2xf(x²)sin[f(x²)]+cos[f(x²)]*[f"(x²)2x*2x+f'(x²)*2]
=-2xf(x²)sin[f(x²)]+4x²f"(x²)cos[f(x²)]+2f'(x²)cos[f(x²)];
(2)y=f²(x)+f(x²),y'=2f(x)*f'(x)+f'(x²)*2x;
y"=[2f'(x)*f'(x)+2f(x)*f"(x)]+f"(x²)*2x*2x+f'(x²)*2=2[f'(x)]²+2f(x)*f"(x)+4x²f"(x²)+2f'(x²);
3、令 u=x^(y²),lnu=y²lnx,等式两端对 x 求导:u'/u=(2y*y')*lnx+y²(1/x),
则 u'=u*[(2y*y')*lnx+y²(1/x)]=2yy'*x^(y²)*lnx+(y²/x)*x^(y²);
原方程两端对 x 求导:[x^(y²)]'+[y²lnx]=0 → [2yy'*x^(y²)*lnx+(y²/x)*x^(y²)]+[2yy'lnx+y²/x]=0;
∴ y'=-[(y²/x)*x^(y²)+(y²/x)]/[2y*x^(y²)*lnx+2ylnx]=-y/(2xlnx);
4、y^(1/x)=x^(1/y),方程两端取对数:(lny)/x=(lnx)/y,即 ylny=xlnx;
再对 x 求导:y'lny+(y/y)y'=lnx+1 → y'=(1+lnx)/(1+lny);
∴ y"=[(1+lnx)/(1+lny)]'=[(1+lny)/x -(1+lnx)*y'/y]/(1+lny)²=[y(1+lny)-xy'(1+lnx)]/[xy(1+lny)²]
=[y(1+lny)²-x(1+lnx)²]/[xy(1+lny)³];
5、y=f(x+y),y'=f'(x+y)*(x+y)'=(1+y')*f'(x+y),∴ y'=f'(x+y)/[1-f'(x+y)];
y"={(1+y')f"(x+y)*[1-f'(x+y)]+f'(x+y)*(1+y')f"(x+y)]} / [1-f'(x+y)]²
={f"(x+y)*[1-f'(x+y)]+f'(x+y)*f"(x+y)}/[1-f'(x+y)]³
=f"(x+y)/[1-f'(x+y)]³;
y"=-sin[f(x²)]f(x²)*2x+cos[f(x²)]*[f'(x²)*2x]'
=-2xf(x²)sin[f(x²)]+cos[f(x²)]*[f"(x²)2x*2x+f'(x²)*2]
=-2xf(x²)sin[f(x²)]+4x²f"(x²)cos[f(x²)]+2f'(x²)cos[f(x²)];
(2)y=f²(x)+f(x²),y'=2f(x)*f'(x)+f'(x²)*2x;
y"=[2f'(x)*f'(x)+2f(x)*f"(x)]+f"(x²)*2x*2x+f'(x²)*2=2[f'(x)]²+2f(x)*f"(x)+4x²f"(x²)+2f'(x²);
3、令 u=x^(y²),lnu=y²lnx,等式两端对 x 求导:u'/u=(2y*y')*lnx+y²(1/x),
则 u'=u*[(2y*y')*lnx+y²(1/x)]=2yy'*x^(y²)*lnx+(y²/x)*x^(y²);
原方程两端对 x 求导:[x^(y²)]'+[y²lnx]=0 → [2yy'*x^(y²)*lnx+(y²/x)*x^(y²)]+[2yy'lnx+y²/x]=0;
∴ y'=-[(y²/x)*x^(y²)+(y²/x)]/[2y*x^(y²)*lnx+2ylnx]=-y/(2xlnx);
4、y^(1/x)=x^(1/y),方程两端取对数:(lny)/x=(lnx)/y,即 ylny=xlnx;
再对 x 求导:y'lny+(y/y)y'=lnx+1 → y'=(1+lnx)/(1+lny);
∴ y"=[(1+lnx)/(1+lny)]'=[(1+lny)/x -(1+lnx)*y'/y]/(1+lny)²=[y(1+lny)-xy'(1+lnx)]/[xy(1+lny)²]
=[y(1+lny)²-x(1+lnx)²]/[xy(1+lny)³];
5、y=f(x+y),y'=f'(x+y)*(x+y)'=(1+y')*f'(x+y),∴ y'=f'(x+y)/[1-f'(x+y)];
y"={(1+y')f"(x+y)*[1-f'(x+y)]+f'(x+y)*(1+y')f"(x+y)]} / [1-f'(x+y)]²
={f"(x+y)*[1-f'(x+y)]+f'(x+y)*f"(x+y)}/[1-f'(x+y)]³
=f"(x+y)/[1-f'(x+y)]³;
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