Question

参数方程的三阶导数公式

Answer 1

x=f(t)
y=g(t)
y'=g'(t)/f'(t)
y''=y'|'t/x'|t
=[g'(t)/f'(t)]'|t//f'(t)
=[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2
y'''=y''|'t/x'|t
={[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2}‘|t//x'|t
={[g''(t)f'(t)-g'(t)f''(t)]'|t*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)+g''(t)f''(t)-g''(t)f''(t)-g'(t)f'''(t)]*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)-g'(t)f'''(t)]*[f'(t)]-2[g''(t)f'(t)-g'(t)f''(t)]f''(t)}/[f'(t)]^4

追答

Answer 2

这里因为d^2y/dx^2=d(y')/dx,
这里y'=dy/dx=g(t)
而因为是参数方程，都要化成对t的求导才行。
所以上式分子分母同时除以dt,
化为：[d(y')/dt]/(dx/dt)
这就是分母里有这个一阶导数的原因。

Answer 3

引用wangwei781999的回答：
x=f(t)
y=g(t)
y'=g'(t)/f'(t)
y''=y'|'t/x'|t
=[g'(t)/f'(t)]'|t//f'(t)
=[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2
y'''=y''|'t/x'|t
={[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2}‘|t//x'|t
={[g''(t)f'(t)-g'(t)f''(t)]'|t*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)+g''(t)f''(t)-g''(t)f''(t)-g'(t)f'''(t)]*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)-g'(t)f'''(t)]*[f'(t)]-2[g''(t)f'(t)-g'(t)f''(t)]f''(t)}/[f'(t)]^4

三阶求导也不对

Answer 4

引用wangwei781999的回答：
x=f(t)
y=g(t)
y'=g'(t)/f'(t)
y''=y'|'t/x'|t
=[g'(t)/f'(t)]'|t//f'(t)
=[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2
y'''=y''|'t/x'|t
={[g''(t)f'(t)-g'(t)f''(t)]/[f'(t)]^2}‘|t//x'|t
={[g''(t)f'(t)-g'(t)f''(t)]'|t*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)+g''(t)f''(t)-g''(t)f''(t)-g'(t)f'''(t)]*[f'(t)]^2-[g''(t)f'(t)-g'(t)f''(t)]*2f'(t)*f''(t)}/[f'(t)]^5
={[g'''(t)f'(t)-g'(t)f'''(t)]*[f'(t)]-2[g''(t)f'(t)-g'(t)f''(t)]f''(t)}/[f'(t)]^4

二阶求导错了