Chain Rule for V=(1/2)*sqrt((v*V)/x)(n*df/dn-f) - Solving for V

[juice34]

Homework Statement

n=y*sqrt((V)/(v*x) and Q=sqrt(v*V*x)*f(n)
so i have V=-dQ/dx=(dQ/dn)*(dn/dx) and the final answer is V=(1/2)*sqrt((v*V)/x)(n*df/dn-f)

Homework Equations

The Attempt at a Solution

i have tried diff. by hand and also by maple and cannot get the answer. What am i doing wrong, because after all i have to quantities added together and that seems weird to me.

 

[Mark44]

What exactly are you trying to do? Your problem statement gives two equations, but doesn't say what you are supposed to do with them or what you are supposed to find.

Also, what's the significance of the underscore on v? You also show it italicized and bolded, which seems like overkill.

 

[juice34]

n and Q are my equations that need to be differentiated. And i need to find V(x)=-dQ/dx=(dQ/dn)*(dn/dx), this differential is using similarity variables. For example when i take V(y)=dQ/dy=(dQ/dn)*(dn/dy), i get sqrt(v*x*V)*(df/dn)*sqrt(V/(v*x)), if you would like me to send you the file if you don't understand still, let me know.
This problem deals with a boundary layer of a flat plate, V= velocity(at a distance infinity away from the plate) v(underscore)=kinematic viscosity, and V(x) and V(y) are the velocity profiles in the x and y direction that need to be found.