第一道the curve with equation y^2=x^3+3x^2 is called the Tschir
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第一道the curve with equation y^2=x^3+3x^2 is called the Tschirnhausen cubic.,find an equation of the tangent line to this curve at point (1,-2).
还有个微分的题
SinX+CosY = sinXcosY
还有个微分的题
SinX+CosY = sinXcosY
先求导
隐函数求导
两边对x求导
2y*y'=3x^2+6x
y'=(3x^2+6x)/(2y)
所以代入x=1,y=-2
y'=(3+6)/(2*(-2))=-9/4
所以切线方程为
(y+2)=(-9/4)(x-1)
y=(-9/4)x+1/4
第二题
SinX+CosY = sinXcosY
两边对x求导,右边积法则
cosx-siny*y'=(sinx)'cosy+sinx(cosy)'
cosx-siny*y'=cosxcosy+sinx(-siny)*y'
(sinxsiny-siny)y'=cosx+cosxcosy
y'=[cosx(1+cosy)]/[siny(sinx-1)]
再问: 我说错了第二题是 是求导, 感谢你啊!等下交作业但这几题做不出
再答: 求了,觉得对就采纳了吧~
隐函数求导
两边对x求导
2y*y'=3x^2+6x
y'=(3x^2+6x)/(2y)
所以代入x=1,y=-2
y'=(3+6)/(2*(-2))=-9/4
所以切线方程为
(y+2)=(-9/4)(x-1)
y=(-9/4)x+1/4
第二题
SinX+CosY = sinXcosY
两边对x求导,右边积法则
cosx-siny*y'=(sinx)'cosy+sinx(cosy)'
cosx-siny*y'=cosxcosy+sinx(-siny)*y'
(sinxsiny-siny)y'=cosx+cosxcosy
y'=[cosx(1+cosy)]/[siny(sinx-1)]
再问: 我说错了第二题是 是求导, 感谢你啊!等下交作业但这几题做不出
再答: 求了,觉得对就采纳了吧~
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