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微积分题目

解:1。∵dy/dx=(xy²-cosxsinx)/(y(1-x²)) ==>y(1-x²)dy=(xy²-cosxsinx)dx ==>y(1-x²)dy-xy²dx+cosxsinxdx=0 ==>(1-x²)d(y²)-y²d(x²)+sin(2x)dx=0 ==>2(1-x²)d(y²)+2y²d(1-x&#...

换元, =∫1/(t+1)d(t²+3)/2 =∫t/(t+1)dt =t-ln(t+1)+C =√(2x-3)-ln(√(2x-3)+1)+C

解:1。∵dy/dx=(xy2-cosxsinx)/(y(1-x2)) ==>y(1-x2)dy=(xy2-cosxsinx)dx ==>y(1-x2)dy-xy2dx+cosxsinxdx=0 ==>(1-x2)d(y2)-y2d(x2)+sin(2x)dx=0 ==>2(1-x2)d(y2)+2y2d(1-x2)+sin(2x)d(2x)=0 ==>2d(y2(1-x2))+sin(2x)d(2x)=0 ==>2y2(1-x2)-cos(2...

1.微分在近似计算中的应用: 要在半径r=1cm的铁球表面上镀一层厚度为0.01cm的铜,求所需铜的重量W(铜的密度k=8.9g/cm^3)(说明:cm^3后面的3是幂,也就是立方厘米,下面的r^3也是指r的3次方,依此类推) 解:先求镀层的体积,再乘以密度,便得...

我们已知 (1)f(x) + f(1-1/x) = 2x, 接下来,用1-1/x代替x写入(1)式,可知 (2)f(1-1/x) + f(1/(1-x)) = 2(1-1/x), 然后,用1/(1-x)代替x写入(1)式,我们有 (3)f(1/(1-x)) + f(x) = 2(1/(1-x)), 通过观察,我们知道(1)(2)(3)...

这种0/0型都可以试试洛必达法则,不是0/0可以变形这个式子变成0/0型再用洛必达。用洛必达的话,这个式子分子分母分别求导:{1/2*1/(4x+1)^1/2}*4➗{1/2*1/(x-1)^1/2}因为x➡️2,所以这个式子的极限=3/4

解:1。∵dy/dx=(xy²-cosxsinx)/(y(1-x²)) ==>y(1-x²)dy=(xy²-cosxsinx)dx ==>y(1-x²)dy-xy²dx+cosxsinxdx=0 ==>(1-x²)d(y²)-y²d(x²)+sin(2x)dx=0 ==>2(1-x²)d(y²)+2y²...

解:1。∵dy/dx=(xy²-cosxsinx)/(y(1-x²)) ==>y(1-x²)dy=(xy²-cosxsinx)dx ==>y(1-x²)dy-xy²dx+cosxsinxdx=0 ==>(1-x²)d(y²)-y²d(x²)+sin(2x)dx=0 ==>2(1-x²)d(y²)+2y²...

必须要用泰勒展开式到四阶才行,因为分母化解的话升到了四阶级,否则永远除以分母为0,望能解答心中的疑问,望采纳哦

解:1。∵dy/dx=(xy²-cosxsinx)/(y(1-x²)) ==>y(1-x²)dy=(xy²-cosxsinx)dx ==>y(1-x²)dy-xy²dx+cosxsinxdx=0 ==>(1-x²)d(y²)-y²d(x²)+sin(2x)dx=0 ==>2(1-x²)d(y²)+2y²...

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