If g is an arbitrary differentiable function of one variable, then
is a solution of
How am I suppose to show this:
Last edited by dwsmith; Dec 25th 2010 at 04:12 PM.
Reason: Adjusted derivative
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There is a mistake. The chain rule
Originally Posted by zzzoak Let
There is a mistake. The chain rule Since g is function of one variable, how can we be sure that the derivative will produce from
For example consider the function
I believe you. I was just asking.
I edited the original post. How can I show g_y=g_x because that is the only way it will be 0?
Please look post #2.
there must be