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设函数y=f(x)由方程sin(xy)+In(y-x)=x确定,求曲线y=f(x)在x=0处的切线方程和法线方程
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设函数y=f(x) 由方程sin(xy)+In(y-x)=x 确定,求曲线y=f(x) 在x=0 处的切线方程和法线方程
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答案和解析
1、确定曲线上的点,将x=0带入原方程,sin(0*y)+ln(y-0)=0,得y=1,即曲线一定点为(0,1);
2、确定切线斜率表达式,即求y’,对原方程两侧求导
cos(xy)*(y+xy')+1/(y-x)*(y'-1)=1,整理,y‘={[y-x]*[1-ycos(xy)]+1}/{[y-x]*xcos(xy)+1}
3、计算给定点(0,1)处切线斜率,y'|x=0=1,即,y'=1
4、点斜式求切线方程:y-1=1*(x-0),即y=x+1
5、法线斜率与切线斜率互为负倒数,y-1=-1*(x-0),即y=-x+1.
2、确定切线斜率表达式,即求y’,对原方程两侧求导
cos(xy)*(y+xy')+1/(y-x)*(y'-1)=1,整理,y‘={[y-x]*[1-ycos(xy)]+1}/{[y-x]*xcos(xy)+1}
3、计算给定点(0,1)处切线斜率,y'|x=0=1,即,y'=1
4、点斜式求切线方程:y-1=1*(x-0),即y=x+1
5、法线斜率与切线斜率互为负倒数,y-1=-1*(x-0),即y=-x+1.
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