l*0
发帖数: 19 | 1 见笑了，很弱的一问题，想不起来怎么做了。
f(x,y), 已知 df/dx, df/dy
s=ax+by, a,b is constant
what is df/ds ? | b**g
发帖数: 335 | 2 chain rule
df/ds = df/dx * (dx/ds) + df/dy * (dy/ds)
【在 l*0 的大作中提到】
: 见笑了，很弱的一问题，想不起来怎么做了。
: f(x,y), 已知 df/dx, df/dy
: s=ax+by, a,b is constant
: what is df/ds ?
| f**********e
发帖数: 97 | 3 我怎么觉得应该是：
(df/dx+df/dy)/(a+b)? | y**k
发帖数: 222 | 4 Interesting...
df/ds=(a*df/dx+b*df/dy)/(a^2+b^2). | e*******6
发帖数: 13 | 5 check Total Derivative on wikipedia~ http://en.wikipedia.org/wiki/Total_derivative
df/ds = df/dx * (dx/ds) + df/dy * (dy/ds) | j******n
发帖数: 271 | 6 I guess you need another variable t=t(x,y) in addition to s. Then find the
jacobian J of the transform. df/ds = first component of J (df/dx df/dy)^T |
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