How Does the Derivative Step Work in Kreyszig's Solution for Y'=(y+9x)^2?

[TRAyres]

This problem is out of Kreyszig's 9E of advanced engineering mathematics.

I don't understand their closed form solution.
Their solution ends up being
y=3tan(3x+c)-9x.

But when they were solving v=y+9x for y,
then taking the derivative, they get:
y'=v'-9, then set that =v^2

--Thats the step I have issues with. They are taking the derivative,
but with respect to what? If it is some dummy variable,
it would be y'=v'-9x' (chain rule?). If it is y with respect to v,
the derivative of y=v-9x would just be y'=v'. Unless x' with respect to
a dummy variable is = 1, of course.

Can someone please enlighten me?

 

[Astronuc]

The problem simply takes dv/dx = d/dx(y+9x) = dy/dx+9 or y'+9 where ' denotes differentiation with respect to x in this case.

 

[TRAyres]

...oh man I was treating y as a constant and not as a variable. y is a function of x, so d/dx(y)=dy/dx...

: smacks forehead : thanks.