Partial derivative - why is it count like this?

player1_1_1
Messages
112
Reaction score
0

Homework Statement


I have John R Taylor "Classical mechanics" part 1, and I have an integral:
\int\limits^{x_2}_{x_1}f\left(y+\alpha\eta,y^{\prime}+\alpha\eta^{\prime},x\right)\mbox{d}x
and here is count derivative of underintegral function in \alpha
\frac{\partial f\left(y+\alpha\eta,y^{\prime}+\alpha\eta^{\prime},x\right)}{\partial\alpha}=\eta\frac{\partial f}{\partial y}+\eta^{\prime}\frac{\partial f}{\partial y^{\prime}}
why there is suddenly \frac{\partial f}{\partial y^{\prime}} and \frac{\partial f}{\partial y}, while this derivative is by \alpha - why not only \eta,\eta^{\prime}?

Homework Equations


I was thinking about function composition derivative, but it didnt helped me.

The Attempt at a Solution


Nothing, I couldn't do anything with this, I don't know why this is count like this, please help;] thanks!
 
Physics news on Phys.org
f=f(y,y',x)

Now if you replace y=\bar{y}+\alpha\eta then y becomes a function of alpha, which means that you have to use the chain rule.

Try choosing an explicit non-linear expression for f if that makes it more clear. For example f(y,y',x) = y2 + y'2
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top