- #1
no_alone
- 32
- 0
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 [tex]f_{x}(0,0) = g_{x}(0,0)[/tex]
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
[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 don't manage to figure how
Thank you
Last edited: