How Do You Prove the Multivariable Chain Rule?

[Lambda96]

Homework Statement

Use the chain rule to show the followin is true $D(f+g)=D(f)+D(g)$

Relevant Equations

Multivariable Chain rule

Hi,

Im completly lost regarding the following exercise:

Unfortunately, I don't understand how to prove the statement using the chain rule. The chain rule is always used if there is a composition, i.e. $f\circ g=f(g(x))$ then I first have to calculate $g(x)$ and insert this result into $f$, but D(f+g) is a operation and not a composition.

 

[fresh_42]

Homework Statement: Use the chain rule to show the followin is true $D(f+g)=D(f)+D(g)$
Relevant Equations: Multivariable Chain rule

Hi,

Im completly lost regarding the following exercise:

View attachment 346915

Unfortunately, I don't understand how to prove the statement using the chain rule. The chain rule is always used if there is a composition, i.e. $f\circ g=f(g(x))$ then I first have to calculate $g(x)$ and insert this result into $f$, but D(f+g) is a operation and not a composition.

The addition is the second function, say $A(f,g)= f+g.$ Then we get $D(f+g)=D(A(f,g))=(D\circ A)(f,g)$ and $D(f)+D(g)= (D(A))(D(f),D(g)).$

 

[Lambda96]

Thank you fresh_42 for your help , now the task makes more sense


If I now apply the chain rule to the expression $(D \circ A)(f,g)$, the result should be $D'(A(f,g))\cdot A'(f,g)$, right?

 

[fresh_42]

I'm not sure how to write it best and it is all about notation.

With $A(f,g)=f+g$ we have
\begin{align*}
A(\alpha (f,g))&=A(\alpha f, \alpha g)=\alpha f+\alpha g=\alpha(f+g)=\alpha A(f,g)\\
A((f_1,g_1)+(f_2,g_2))&=A((f_1+f_2),(g_1+g_2))=(f_1+f_2)+(g_1+g_2)\\
&=(f_1+g_1)+(f_2+g_2)=A(f_1,g_1)+A(f_2,g_2)
\end{align*}
and $A$ is linear. Therefore we have $DA=A.$ Finally,
$$
D(f+g)=D(A(f,g))=DA(D(f,g))=DA(D(f),D(g))=A(D(f),D(g))=D(f)+D(g)
$$

Please note to every equation sign which property we have used!
1. ...
2. ...
3. ...
4. ...
5. ...

 

[Lambda96]

Many thanks for your help fresh_42 now I have understood it