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sgh1324

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df=∂f/∂x dx+∂f/∂y dy and ∂f/∂x=p,∂f/∂y=q

So we get

df=p dx+q dy

d(f−qy)=p dx−y dqand now, define g.

g=f−q y

dg = p dx - y dq

and then I faced problem.

∂g/∂x=p←←←←←←←←←←←←←←← book said like this because we can see g is a function of x and p so that chain rule makes ∂g/∂x=p

but I wrote directly ∂g/∂x so i can get a result,

∂g/∂x=∂f/∂x−y ∂q/∂x−q ∂y∂x=p−y ∂q∂x - 0(y is independent of x)

I know that variation y is independent of x, but I'm not sure that q is also the independent function of x. what if the func f is xy? if I set like that,

g=f−qy=xy−xy=0

it's not same with p!mathematical methods in the physical sciences, Mary L. Boas 3rd edition page231

http://www.utdallas.edu/~pervin/ENGR3300/Boaz.pdf