Derivative of a composition of differentiable function and a continously differentiable curve

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Let $f:mathbbR^2tomathbbR$ be a differential function at the origin, and $gamma_1,gamma_2: (-1,1)to mathbbR^2$ continously differentiable curves at the origin, s.t. $gamma_1(0)=gamma_2(0)=(0,0)$, and $forall tin(-1,1) :gamma_2(t)=gamma_1(t)+(t^2,t^3)$.

Show $frac d (f(gamma_1(t)) dtbiggrvert_t=0 = frac d (f(gamma_2(t)) dt biggrvert_t=0$.

I tried applying the chain rule the following way:

1)$D_fcircgamma_1(t)(0)=D_f(0)D_gamma_1(0)$.

2)
$D_fcircgamma_2(t)(0)=D_f(0)D_gamma_2(0)=D_f(0)D_gamma_1+(t^2,t^3)(0)=^*D_f(0)(D_gamma_1+(2t,3t^2)(0)=D_f(0)D_gamma_1(0)$

(*)- From linearity of the derivative operation

Is this use of the chain rule correct? And if so, where did I use continuity of the derivatives of $gamma$?

multivariable-calculus derivatives

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asked Jul 23 at 20:37

Sar

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Let $f:mathbbR^2tomathbbR$ be a differential function at the origin, and $gamma_1,gamma_2: (-1,1)to mathbbR^2$ continously differentiable curves at the origin, s.t. $gamma_1(0)=gamma_2(0)=(0,0)$, and $forall tin(-1,1) :gamma_2(t)=gamma_1(t)+(t^2,t^3)$.

Show $frac d (f(gamma_1(t)) dtbiggrvert_t=0 = frac d (f(gamma_2(t)) dt biggrvert_t=0$.

I tried applying the chain rule the following way:

1)$D_fcircgamma_1(t)(0)=D_f(0)D_gamma_1(0)$.

2)
$D_fcircgamma_2(t)(0)=D_f(0)D_gamma_2(0)=D_f(0)D_gamma_1+(t^2,t^3)(0)=^*D_f(0)(D_gamma_1+(2t,3t^2)(0)=D_f(0)D_gamma_1(0)$

(*)- From linearity of the derivative operation

Is this use of the chain rule correct? And if so, where did I use continuity of the derivatives of $gamma$?

multivariable-calculus derivatives

share|cite|improve this question

asked Jul 23 at 20:37

Sar

40410

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $f:mathbbR^2tomathbbR$ be a differential function at the origin, and $gamma_1,gamma_2: (-1,1)to mathbbR^2$ continously differentiable curves at the origin, s.t. $gamma_1(0)=gamma_2(0)=(0,0)$, and $forall tin(-1,1) :gamma_2(t)=gamma_1(t)+(t^2,t^3)$.

Show $frac d (f(gamma_1(t)) dtbiggrvert_t=0 = frac d (f(gamma_2(t)) dt biggrvert_t=0$.

I tried applying the chain rule the following way:

1)$D_fcircgamma_1(t)(0)=D_f(0)D_gamma_1(0)$.

2)
$D_fcircgamma_2(t)(0)=D_f(0)D_gamma_2(0)=D_f(0)D_gamma_1+(t^2,t^3)(0)=^*D_f(0)(D_gamma_1+(2t,3t^2)(0)=D_f(0)D_gamma_1(0)$

(*)- From linearity of the derivative operation

Is this use of the chain rule correct? And if so, where did I use continuity of the derivatives of $gamma$?

multivariable-calculus derivatives

share|cite|improve this question

asked Jul 23 at 20:37

Sar

40410

Let $f:mathbbR^2tomathbbR$ be a differential function at the origin, and $gamma_1,gamma_2: (-1,1)to mathbbR^2$ continously differentiable curves at the origin, s.t. $gamma_1(0)=gamma_2(0)=(0,0)$, and $forall tin(-1,1) :gamma_2(t)=gamma_1(t)+(t^2,t^3)$.

Show $frac d (f(gamma_1(t)) dtbiggrvert_t=0 = frac d (f(gamma_2(t)) dt biggrvert_t=0$.

I tried applying the chain rule the following way:

1)$D_fcircgamma_1(t)(0)=D_f(0)D_gamma_1(0)$.

2)
$D_fcircgamma_2(t)(0)=D_f(0)D_gamma_2(0)=D_f(0)D_gamma_1+(t^2,t^3)(0)=^*D_f(0)(D_gamma_1+(2t,3t^2)(0)=D_f(0)D_gamma_1(0)$

(*)- From linearity of the derivative operation

Is this use of the chain rule correct? And if so, where did I use continuity of the derivatives of $gamma$?

multivariable-calculus derivatives

share|cite|improve this question

asked Jul 23 at 20:37

Sar

40410

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 20:37

Sar

40410

asked Jul 23 at 20:37

Sar

40410

asked Jul 23 at 20:37

Sar

40410

40410

add a comment | 

add a comment | 

1 Answer
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up vote
1
down vote

It's correct. A simpler argument would be to note that $D_gamma_1(0)=D_gamma_2(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true even if γ are differentiable with non-continuous derivative ?
  – Sar
  Jul 24 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
  – Christian Blatter
  Jul 24 at 12:19
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

It's correct. A simpler argument would be to note that $D_gamma_1(0)=D_gamma_2(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true even if γ are differentiable with non-continuous derivative ?
  – Sar
  Jul 24 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
  – Christian Blatter
  Jul 24 at 12:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

It's correct. A simpler argument would be to note that $D_gamma_1(0)=D_gamma_2(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true even if γ are differentiable with non-continuous derivative ?
  – Sar
  Jul 24 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
  – Christian Blatter
  Jul 24 at 12:19
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

It's correct. A simpler argument would be to note that $D_gamma_1(0)=D_gamma_2(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

It's correct. A simpler argument would be to note that $D_gamma_1(0)=D_gamma_2(0)$. This, together with a pointer to the chain rule, proves the claim without any calculations.

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 9:19

Christian Blatter

163k7107306

answered Jul 24 at 9:19

Christian Blatter

163k7107306

answered Jul 24 at 9:19

Christian Blatter

163k7107306

163k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true even if γ are differentiable with non-continuous derivative ?
  – Sar
  Jul 24 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
  – Christian Blatter
  Jul 24 at 12:19
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So it is true even if γ are differentiable with non-continuous derivative ?
  – Sar
  Jul 24 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
  – Christian Blatter
  Jul 24 at 12:19
  

  
  
  
  

So it is true even if γ are differentiable with non-continuous derivative ?
– Sar
Jul 24 at 11:04

So it is true even if γ are differentiable with non-continuous derivative ?
– Sar
Jul 24 at 11:04

The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
– Christian Blatter
Jul 24 at 12:19

The chain rule does not assume the continuity of the derivative. On the other hand in most cases the derivative is continuous anyway.
– Christian Blatter
Jul 24 at 12:19

add a comment | 

 

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