Prove differentiable for a function

[no_alone]

Homework Statement

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 $$f_{x}(0,0) = g_{x}(0,0)$$

Homework Equations

all calculus

The Attempt at a Solution

Well my idea is like this
First I know that because f(x,y) differentiable at (0,0) so
$$f(0+x,0+y)= f(0,0) +f_{x}(0,0)*x + f_{y}(0,0)*y + o(||x,y||)$$

$$lim_{(x,y)->(0,0)} \frac{f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0$$

and when trying to prove that g(x,y) is differentiable I need to prove that
$$g(0+x,0+y) = g(0,0) + g_{x}(0,0)*x + g_{y}(0,0)*y + o(||x,y||)$$
$$g(0+x,0+y) = 0 + f_{x}(0,0)*q'(0)*x + f_{y}(0,0)*q'(0)*y + o(||x,y||)$$

$$g(0+x,0+y) = 0 + f_{x}(0,0)*1*x + f_{y}(0,0)*1*y + o(||x,y||)$$
$$lim_{(x,y)->(0,0)} \frac{q(f(x,y))-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0$$
And I'm a bit stuck in here..
logically I think I can use the fact that q'(0) = 1
$$lim_{(x)->(0)} \frac{q(0+x)+q(0)}{x} =\frac{q(x)}{x}= 1$$
But I don't manage to figure how

Thank you

 

[no_alone]

Anyone?
I've made some more progress but its not enough,see
Because I know that q(x) has a derivative in t=0 then it means that
q(0+x) = q(0)+q'(0)x+o(|x|) (x->0)
Thats mean that: [ q(0)= 0 q'(0)=1)
$$lim_{x->0}\frac{q(x)-x}{|x|}=0$$

And I think maybe I need to use this with the knowledge that

$$lim_{(x,y)->(0,0)} \frac{f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y}{||x,y||} = 0$$

Meaning that
$$f(x,y)-f_{x}(0,0)*x - f_{y}(0,0)*y = v(x,y)\sqrt{x^{2}+y^{2}}$$
while $$lim_{(x.y)->(0,0)}v(x,y)=0$$
And that
q(f(x,y)) = g(x,y)

But I don't know how

Anyone Please?