有关z=arctan(y/x)的一阶偏导数z=arctan(y/x)的一阶偏导数,你给的答案是 ∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)= -y/(x²+y²)∂z/∂y= {1/[1+(y/x)²]}/x= x/(x²+y²)我想问下,arctanX的导
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![有关z=arctan(y/x)的一阶偏导数z=arctan(y/x)的一阶偏导数,你给的答案是 ∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)= -y/(x²+y²)∂z/∂y= {1/[1+(y/x)²]}/x= x/(x²+y²)我想问下,arctanX的导](/uploads/image/z/7631142-6-2.jpg?t=%E6%9C%89%E5%85%B3z%3Darctan%28y%2Fx%29%E7%9A%84%E4%B8%80%E9%98%B6%E5%81%8F%E5%AF%BC%E6%95%B0z%3Darctan%28y%2Fx%29%E7%9A%84%E4%B8%80%E9%98%B6%E5%81%8F%E5%AF%BC%E6%95%B0%2C%E4%BD%A0%E7%BB%99%E7%9A%84%E7%AD%94%E6%A1%88%E6%98%AF+%26%238706%3Bz%2F%26%238706%3Bx%3D+%7B1%2F%5B1%2B%28y%2Fx%29%26%23178%3B%5D%7D%C2%B7%28-y%2Fx%26%23178%3B%29%3D+-y%2F%28x%26%23178%3B%2By%26%23178%3B%29%26%238706%3Bz%2F%26%238706%3By%3D+%7B1%2F%5B1%2B%28y%2Fx%29%26%23178%3B%5D%7D%2Fx%3D+x%2F%28x%26%23178%3B%2By%26%23178%3B%29%E6%88%91%E6%83%B3%E9%97%AE%E4%B8%8B%2CarctanX%E7%9A%84%E5%AF%BC)
有关z=arctan(y/x)的一阶偏导数z=arctan(y/x)的一阶偏导数,你给的答案是 ∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)= -y/(x²+y²)∂z/∂y= {1/[1+(y/x)²]}/x= x/(x²+y²)我想问下,arctanX的导
有关z=arctan(y/x)的一阶偏导数
z=arctan(y/x)的一阶偏导数,你给的答案是
∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)
= -y/(x²+y²)
∂z/∂y= {1/[1+(y/x)²]}/x
= x/(x²+y²)
我想问下,arctanX的导数公式不是1/(1+x²)吗,你∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)中,大括号右边的(-y/x²)部分是怎么来的?∂z/∂y= {1/[1+(y/x)²]}/x同样为什么要除以x,请详细指教下,
有关z=arctan(y/x)的一阶偏导数z=arctan(y/x)的一阶偏导数,你给的答案是 ∂z/∂x= {1/[1+(y/x)²]}·(-y/x²)= -y/(x²+y²)∂z/∂y= {1/[1+(y/x)²]}/x= x/(x²+y²)我想问下,arctanX的导
∂z/∂x= {1/[1+(y/x)²]}(y/x)`= {1/[1+(y/x)²]}(-y/x²) (这是复合函数求导,即要对(y/x)中的x求导,
即(y/x)`=-y/x²)是这样来的
同样∂z/∂y= {1/[1+(y/x)²]}(y/x)`= {1/[1+(y/x)²]}/x (即要对(y/x)中的y求导,即(y/x)`=1/x)
这个是复合函数求导的,你可以先令y/x=t,这样就知道原因了