# How Do You Prove the Multivariable Chain Rule?

• Lambda96
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.

Lambda96 said:
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?

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. ...

Last edited:
Lambda96
Many thanks for your help fresh_42 now I have understood it

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