GCT said:
\frac{d(a^{-xy})}{dx}=a^{xy}a^{-xy}\frac{d(-xy)}{d(xy)}\log(a)
how'd you take a derivative of two variables, is this multivariable calculus?
It is just the chain rule of single variable calculus.
\frac{df}{dx}=\frac{df}{du} \ \frac{du}{dx}
So we can differentiate with respect to xy or some other function of x. What ever function we use is u in the above. For example
\frac{d(\sin(xy))}{d(xy)}=\cos(xy)
The xy can be treated as a variable. It might be easier to see when xy is written u=xy.
The equation you quote is only true of y that satisfy the differential equation, but it is easy to see its truth in those cases.
\frac{d(a^{-xy})}{dx}=-a^{-xy}\log(a)\frac{d(xy)}{dx}=-a^{xy}a^{-xy}\log(a)=-\log(a)
where the given differential equation
\frac{d(xy)}{dx}=a^{xy}
has been used
I will do this once more, attempting to avoid writing derivatives that look strange. All three of these workings are essentially the same.
The given equation
\frac{d(xy)}{dx}=a^{ax}
obvious algebra
a^{-xy}\frac{d(xy)}{dx}=1
integrate
\int a^{-xy}\frac{d(xy)}{dx} \ dx=\int 1 \ dx
performing the integration (the left can be easily seen substituting u=xy)
\frac{-1}{\log(a)}a^{-xy}=x+c
now the same letting u=xy throughout
\frac{d(u)}{dx}=a^{u}
obvious algebra
a^{-u}\frac{d(u)}{dx}=1
integrate
\int a^{-u}\frac{d(u)}{dx} \ dx=\int 1 \ dx
performing the integration
\frac{-1}{\log(a)}a^{-u}=x+c
