(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

f(x,y) is differentiable in (0,0) and f(0,0)=0

q(t) devirate t=0 and q'(0)=1 q(0)=0

let it be g(x,y) = q(f(x,y))

prove that g differentiable in (0,0) and that [tex]f_{x}(0,0) = g_{x}(0,0)[/tex]

2. Relevant equations

all calculus

3. The attempt at a solution

Well my idea is like this

First I know that because f(x,y) differentiable at (0,0) so

[tex]f(0+x,0+y)= f(0,0) +f_{x}(0,0)*x + f_{y}(0,0)*y + o(||x,y||)[/tex]

[tex] lim_{(x,y)->(0,0)} \frac{f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0[/tex]

and when trying to prove that g(x,y) is differentiable I need to prove that

[tex]g(0+x,0+y) = g(0,0) + g_{x}(0,0)*x + g_{y}(0,0)*y + o(||x,y||)[/tex]

[tex]g(0+x,0+y) = 0 + f_{x}(0,0)*q'(0)*x + f_{y}(0,0)*q'(0)*y + o(||x,y||)[/tex]

[tex]g(0+x,0+y) = 0 + f_{x}(0,0)*1*x + f_{y}(0,0)*1*y + o(||x,y||)[/tex]

[tex] lim_{(x,y)->(0,0)} \frac{q(f(x,y))-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0[/tex]

And I'm a bit stuck in here..

logically I think I can use the fact that q'(0) = 1

[tex] lim_{(x)->(0)} \frac{q(0+x)+q(0)}{x} =\frac{q(x)}{x}= 1[/tex]

But I dont manage to figure how

Thank you

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# Homework Help: Prove differentiable for a function

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