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Finding the partial derivative from the given information

  • Thread starter Amadeo
  • Start date
Problem Statement
see post
Relevant Equations
dz/dr = dz/dx (dx/dr) + dz/dy(dy/dr)
problem13.PNG


It seems that the way to combine the information given is

z = f ( g ( (3r^3 - s^2), (re^s) ) )

we know that the multi-variable chain rule is

(dz/dr) = (dz/dx)* dx/dr + (dz/dy)*dy/dr

and

(dz/ds) = (dz/dx)* dx/ds + (dz/dy)*dy/ds

---(Parentheses indicate partial derivative)

other perhaps useful information

(dx/dr)= 9r^2
(dx/ds)=-2s
(dy/dr)=e^s
(dy/ds)=re^s

I don't know how to apply this information because, usually, there is only one variable, like t, being fed into the multi-input function, and the chain rule works nicely. But here we have r and s being fed into the multi-input function g. Further, g is the input of the function f, which is z. In order to obtain the derivative of z with respect to g, we would need to know the function f(g). It is not given. I would guess that the way to find the derivative of z with respect to r would be to multiply the derivative of z with respect to g by the derivative of g with respect to x, multiplied by the derivative of x with respect to r, plus the derivative of z with respect to g multiplied by the derivative of g with respect to y, multiplied by the derivative of y with respect to r.

So that is where I am on this. Thanks for any assistance.
 

haruspex

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here we have r and s being fed into the multi-input function g.
So apply the Relevant Equation you listed, with g in place of z.
g is the input of the function f, which is z. In order to obtain the derivative of z with respect to g, we would need to know the function f(g).
No, you only need to know the appropriate derivatives at the point of interest. You are given some values for gx, gy and f' (i.e., fg).
 
Thank you. got it.

42, -24.
 

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