MHB Solving a Chain Rule Problem with F(x,y)

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To solve the chain rule problem for F(x,y) = f((x-y)/(x+y)), the partial derivatives need to be calculated using the chain rule. The first derivative ∂F/∂x at the point (2,-1) is determined to be -80. The second mixed derivative ∂²F/∂x∂y is calculated to be 372. The discussion emphasizes the importance of correctly applying the chain rule and evaluating derivatives at the specified point. Understanding the relationships between the variables t and s is crucial for simplifying the calculations.
Yankel
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Hello all,

I need some help with this chain rule problem.

\[F(x,y)=f\left (\frac{x-y}{x+y} \right )\]

It is known that:

f'(1)=20,f'(2)=30, f'(3)=40

and

\[f''(1)=5,f''(2)=6,f''(3))=7\]Find

\frac{\partial F}{\partial x}(2,-1)

and

\[\frac{\partial^2 F}{\partial x\partial y}\]The final answers should be -80 and 372.

I am less bothered with the final numbers (although would like to get there). It is the way that I am interested in. I was thinking to set t=x-y and s=x+y, but it got me stuck.
 
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Are both derivatives being evaluated at $(2,-1)$?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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