Question

how to prove that (1/(f(D_x,D_y))*e^(ax+by))=(1/f(a,b))*)e^(ax+by)? if q(x,y) exponential.

Answer 1

I assume you're question is:

\[\frac{1}{f(Dx, Dy)} * e^(ax+by) = \frac{ 1 }{ f(a,b)} * e^(ax + by)\]

The equation is rather vague, but if you take e^(ax+by) alone, its derivative is \[d(e^(ax+by)) = (adx + bdy)e^(ax+by)\]
The q(x,y) isn't even in you're question. Is there more to this?

Answer 2

q(x,y)=e^ax+by

Answer 3

youre' right polaris,how to prove\[\frac{ 1 }{ f(Dx,Dy) }*e ^{ax+by}=\frac{ 1 }{ f(a,b) }*e ^{ax+by}\]

Answer 4

so is the Dx = Dx(q) and Dy = Dy(q)?

Answer 5

because you equation is correct if (ax + by) is the function you are taking the derivative of.

Answer 6

derivative of?

Answer 7

Dx is partial Derivative of X correct?
Partial of (ax + by) with respect to X is a
partial of (ax + by) with respect to Y is b,

Or am I way off of what this question is asking?

Answer 8

been learning more latex so I can type this properly:

\[\frac{1}{f(\partial_x, \partial_y)}e^{ax+by} = \frac{1}{f(a,b)}e^{ax+by}\]

so what I'm saying is that if
\[P = (ax + by)\]
\[P_x = \frac{\partial}{\partial x}(ax+by) = a\]
\[P_y = \frac{\partial}{\partial y}(ax+by) = b\]
\[f(P_x,P_y) = f(a,b)\]

I'm just not sure this is what you are looking for or not, because it seems way to obvious.

Answer 9

i think you has been help me