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Inner product of differentiable functions
Clash Royale CLAN TAG#URR8PPP
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Suppose that f,g are differentiable functions from $mathcalR$ to $mathcalR^n$.
Show that $langle f(t),g(t)rangle' = langle f'(t),g(t)rangle + langle f(t),g'(t)rangle$
I've been banging my head against a brick wall on this for a while, I can see that the RHS is something like the product rule for differentiation. We're not given an inner product but even just trying to by parts integrate the standard inner wasn't working.
Thanks in advance for any help.
linear-algebra inner-product-space
edited 1 hour ago
edm
3,2931425
asked 1 hour ago
Milan Leonard
909
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up vote
1
down vote
favorite
Suppose that f,g are differentiable functions from $mathcalR$ to $mathcalR^n$.
Show that $langle f(t),g(t)rangle' = langle f'(t),g(t)rangle + langle f(t),g'(t)rangle$
I've been banging my head against a brick wall on this for a while, I can see that the RHS is something like the product rule for differentiation. We're not given an inner product but even just trying to by parts integrate the standard inner wasn't working.
Thanks in advance for any help.
linear-algebra inner-product-space
edited 1 hour ago
edm
3,2931425
asked 1 hour ago
Milan Leonard
909
Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
1
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that f,g are differentiable functions from $mathcalR$ to $mathcalR^n$.
Show that $langle f(t),g(t)rangle' = langle f'(t),g(t)rangle + langle f(t),g'(t)rangle$
I've been banging my head against a brick wall on this for a while, I can see that the RHS is something like the product rule for differentiation. We're not given an inner product but even just trying to by parts integrate the standard inner wasn't working.
Thanks in advance for any help.
linear-algebra inner-product-space
edited 1 hour ago
edm
3,2931425
asked 1 hour ago
Milan Leonard
909
Suppose that f,g are differentiable functions from $mathcalR$ to $mathcalR^n$.
Show that $langle f(t),g(t)rangle' = langle f'(t),g(t)rangle + langle f(t),g'(t)rangle$
I've been banging my head against a brick wall on this for a while, I can see that the RHS is something like the product rule for differentiation. We're not given an inner product but even just trying to by parts integrate the standard inner wasn't working.
Thanks in advance for any help.
linear-algebra inner-product-space
linear-algebra inner-product-space
edited 1 hour ago
edm
3,2931425
asked 1 hour ago
Milan Leonard
909
edited 1 hour ago
edm
3,2931425
asked 1 hour ago
Milan Leonard
909
edited 1 hour ago
edm
3,2931425
edited 1 hour ago
edm
3,2931425
edited 1 hour ago
edm
3,2931425
3,2931425
asked 1 hour ago
Milan Leonard
909
asked 1 hour ago
Milan Leonard
909
asked 1 hour ago
Milan Leonard
909
909
Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
1
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago
add a comment |Â
Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
1
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago
Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
1
1
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,gcolon Bbb RtoBbb R$. To wit, write
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fraclangle f(t+h),g(t+h)rangle-langle f(t),g(t)rangleh.
endalign*
Let me write $fg$, i.e. just the concatenation, to mean $langle f,grangle$. There shouldn't be any confusion by doing this. With this notation, we have
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fracf(t+h)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)-f(t)hg(t+h) + f(t)lim_hto0fracg(t+h)-g(t)h\
&= f'(t)g(t) + f(t)g'(t) \
&= langle f'(t),g(t)rangle + langle f(t),g'(t)rangle,
endalign*
where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $langlecdot,cdotranglecolon Bbb R^ntimesBbb R^ntoBbb R$.
answered 1 hour ago
AOrtiz
10.1k21239
add a comment |Â
up vote
2
down vote
Write $f(t)=(f_1(t),f_2(t),dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),dots,g_n(t))$. The inner product is given by $$langle f(t),g(t)rangle=sum_i=1^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$fracddtlangle f(t),g(t)rangle=sum_i=1^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$langle f'(t),g(t)rangle + langle f(t),g'(t)rangle.$$
There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$
Your map $tmapstolangle f(t),g(t)rangle$ is the composition $tmapsto(f(t),g(t))$, $(x,y)mapstolangle x,yrangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.
answered 1 hour ago
edm
3,2931425
add a comment |Â
up vote
0
down vote
A slightly different approach.
Let $fin C^1(Bbb R^ntimes Bbb R^m,Bbb R^ell)$, then it is easy to check that
$$partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)btag1$$
where $partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,cdot)$.
Let $d:Bbb R^ntimes Bbb R^ntoBbb R$ a dot product, hence
$$partial, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)btag2$$
Now note that the functions $d(cdot ,y)$ and $d(x,cdot)$ are linear, so
$$D_x d(x,y)=d(cdot, y)implies D_x d(x,y)a=d(a,y)tag3$$
And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that
$$partial, d(x,y)(a,b)=d(a,y)+d(x,b)tag4$$
By last, using the chain rule, we have that
$$partial, [d(f,g)]=partial d(f,g)partial(f,g)=partial d(f,g)(f',g')\=d(f',g)+d(f,g')tag5$$
The key points here are $(1)$ and the fact that if $A$ is a linear function then $partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.
answered 1 hour ago
Masacroso
11.9k41745
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,gcolon Bbb RtoBbb R$. To wit, write
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fraclangle f(t+h),g(t+h)rangle-langle f(t),g(t)rangleh.
endalign*
Let me write $fg$, i.e. just the concatenation, to mean $langle f,grangle$. There shouldn't be any confusion by doing this. With this notation, we have
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fracf(t+h)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)-f(t)hg(t+h) + f(t)lim_hto0fracg(t+h)-g(t)h\
&= f'(t)g(t) + f(t)g'(t) \
&= langle f'(t),g(t)rangle + langle f(t),g'(t)rangle,
endalign*
where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $langlecdot,cdotranglecolon Bbb R^ntimesBbb R^ntoBbb R$.
answered 1 hour ago
AOrtiz
10.1k21239
add a comment |Â
up vote
2
down vote
accepted
The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,gcolon Bbb RtoBbb R$. To wit, write
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fraclangle f(t+h),g(t+h)rangle-langle f(t),g(t)rangleh.
endalign*
Let me write $fg$, i.e. just the concatenation, to mean $langle f,grangle$. There shouldn't be any confusion by doing this. With this notation, we have
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fracf(t+h)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)-f(t)hg(t+h) + f(t)lim_hto0fracg(t+h)-g(t)h\
&= f'(t)g(t) + f(t)g'(t) \
&= langle f'(t),g(t)rangle + langle f(t),g'(t)rangle,
endalign*
where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $langlecdot,cdotranglecolon Bbb R^ntimesBbb R^ntoBbb R$.
answered 1 hour ago
AOrtiz
10.1k21239
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,gcolon Bbb RtoBbb R$. To wit, write
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fraclangle f(t+h),g(t+h)rangle-langle f(t),g(t)rangleh.
endalign*
Let me write $fg$, i.e. just the concatenation, to mean $langle f,grangle$. There shouldn't be any confusion by doing this. With this notation, we have
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fracf(t+h)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)-f(t)hg(t+h) + f(t)lim_hto0fracg(t+h)-g(t)h\
&= f'(t)g(t) + f(t)g'(t) \
&= langle f'(t),g(t)rangle + langle f(t),g'(t)rangle,
endalign*
where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $langlecdot,cdotranglecolon Bbb R^ntimesBbb R^ntoBbb R$.
answered 1 hour ago
AOrtiz
10.1k21239
The proof is very similar to how you prove that $(fg)' = f'g + fg'$ for any functions $f,gcolon Bbb RtoBbb R$. To wit, write
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fraclangle f(t+h),g(t+h)rangle-langle f(t),g(t)rangleh.
endalign*
Let me write $fg$, i.e. just the concatenation, to mean $langle f,grangle$. There shouldn't be any confusion by doing this. With this notation, we have
beginalign*
langle f(t),g(t)rangle' &= lim_hto0fracf(t+h)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)g(t+h)-f(t)g(t+h)+f(t)g(t+h)-f(t)g(t)h \
&= lim_hto0fracf(t+h)-f(t)hg(t+h) + f(t)lim_hto0fracg(t+h)-g(t)h\
&= f'(t)g(t) + f(t)g'(t) \
&= langle f'(t),g(t)rangle + langle f(t),g'(t)rangle,
endalign*
where we switched back to the usual notation in the last line. In the second-to-last equality, we used continuity of the inner product $langlecdot,cdotranglecolon Bbb R^ntimesBbb R^ntoBbb R$.
answered 1 hour ago
AOrtiz
10.1k21239
answered 1 hour ago
AOrtiz
10.1k21239
answered 1 hour ago
AOrtiz
10.1k21239
answered 1 hour ago
AOrtiz
10.1k21239
10.1k21239
add a comment |Â
add a comment |Â
up vote
2
down vote
Write $f(t)=(f_1(t),f_2(t),dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),dots,g_n(t))$. The inner product is given by $$langle f(t),g(t)rangle=sum_i=1^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$fracddtlangle f(t),g(t)rangle=sum_i=1^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$langle f'(t),g(t)rangle + langle f(t),g'(t)rangle.$$
There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$
Your map $tmapstolangle f(t),g(t)rangle$ is the composition $tmapsto(f(t),g(t))$, $(x,y)mapstolangle x,yrangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.
answered 1 hour ago
edm
3,2931425
add a comment |Â
up vote
2
down vote
Write $f(t)=(f_1(t),f_2(t),dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),dots,g_n(t))$. The inner product is given by $$langle f(t),g(t)rangle=sum_i=1^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$fracddtlangle f(t),g(t)rangle=sum_i=1^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$langle f'(t),g(t)rangle + langle f(t),g'(t)rangle.$$
There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$
Your map $tmapstolangle f(t),g(t)rangle$ is the composition $tmapsto(f(t),g(t))$, $(x,y)mapstolangle x,yrangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.
answered 1 hour ago
edm
3,2931425
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Write $f(t)=(f_1(t),f_2(t),dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),dots,g_n(t))$. The inner product is given by $$langle f(t),g(t)rangle=sum_i=1^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$fracddtlangle f(t),g(t)rangle=sum_i=1^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$langle f'(t),g(t)rangle + langle f(t),g'(t)rangle.$$
There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$
Your map $tmapstolangle f(t),g(t)rangle$ is the composition $tmapsto(f(t),g(t))$, $(x,y)mapstolangle x,yrangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.
answered 1 hour ago
edm
3,2931425
Write $f(t)=(f_1(t),f_2(t),dots,f_n(t))$ and $g(t)=(g_1(t),g_2(t),dots,g_n(t))$. The inner product is given by $$langle f(t),g(t)rangle=sum_i=1^nf_i(t)g_i(t).$$ You apply sum rule and product rule of differentiation to see that $$fracddtlangle f(t),g(t)rangle=sum_i=1^nf_i'(t)g_i(t)+f_i(t)g_i'(t).$$ The last expression is equal to $$langle f'(t),g(t)rangle + langle f(t),g'(t)rangle.$$
There is also a general formula telling us how to calculate derivatives of bilinear functions, but you may not yet be ready to understand. Given a bilinear function $B$ that takes a pair of vectors $(x,y)$ as input, the derivative at $(a,b)$ is the linear map $B'(a,b)$ that acts on $(x,y)$ by the formula $$B'(a,b)(x,y)=B(a,y)+B(x,b).$$
Your map $tmapstolangle f(t),g(t)rangle$ is the composition $tmapsto(f(t),g(t))$, $(x,y)mapstolangle x,yrangle$. So alternatively, you can apply chain rule to obtain the formula for its derivative.
answered 1 hour ago
edm
3,2931425
answered 1 hour ago
edm
3,2931425
answered 1 hour ago
edm
3,2931425
answered 1 hour ago
edm
3,2931425
3,2931425
add a comment |Â
add a comment |Â
up vote
0
down vote
A slightly different approach.
Let $fin C^1(Bbb R^ntimes Bbb R^m,Bbb R^ell)$, then it is easy to check that
$$partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)btag1$$
where $partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,cdot)$.
Let $d:Bbb R^ntimes Bbb R^ntoBbb R$ a dot product, hence
$$partial, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)btag2$$
Now note that the functions $d(cdot ,y)$ and $d(x,cdot)$ are linear, so
$$D_x d(x,y)=d(cdot, y)implies D_x d(x,y)a=d(a,y)tag3$$
And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that
$$partial, d(x,y)(a,b)=d(a,y)+d(x,b)tag4$$
By last, using the chain rule, we have that
$$partial, [d(f,g)]=partial d(f,g)partial(f,g)=partial d(f,g)(f',g')\=d(f',g)+d(f,g')tag5$$
The key points here are $(1)$ and the fact that if $A$ is a linear function then $partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.
answered 1 hour ago
Masacroso
11.9k41745
add a comment |Â
up vote
0
down vote
A slightly different approach.
Let $fin C^1(Bbb R^ntimes Bbb R^m,Bbb R^ell)$, then it is easy to check that
$$partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)btag1$$
where $partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,cdot)$.
Let $d:Bbb R^ntimes Bbb R^ntoBbb R$ a dot product, hence
$$partial, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)btag2$$
Now note that the functions $d(cdot ,y)$ and $d(x,cdot)$ are linear, so
$$D_x d(x,y)=d(cdot, y)implies D_x d(x,y)a=d(a,y)tag3$$
And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that
$$partial, d(x,y)(a,b)=d(a,y)+d(x,b)tag4$$
By last, using the chain rule, we have that
$$partial, [d(f,g)]=partial d(f,g)partial(f,g)=partial d(f,g)(f',g')\=d(f',g)+d(f,g')tag5$$
The key points here are $(1)$ and the fact that if $A$ is a linear function then $partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.
answered 1 hour ago
Masacroso
11.9k41745
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A slightly different approach.
Let $fin C^1(Bbb R^ntimes Bbb R^m,Bbb R^ell)$, then it is easy to check that
$$partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)btag1$$
where $partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,cdot)$.
Let $d:Bbb R^ntimes Bbb R^ntoBbb R$ a dot product, hence
$$partial, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)btag2$$
Now note that the functions $d(cdot ,y)$ and $d(x,cdot)$ are linear, so
$$D_x d(x,y)=d(cdot, y)implies D_x d(x,y)a=d(a,y)tag3$$
And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that
$$partial, d(x,y)(a,b)=d(a,y)+d(x,b)tag4$$
By last, using the chain rule, we have that
$$partial, [d(f,g)]=partial d(f,g)partial(f,g)=partial d(f,g)(f',g')\=d(f',g)+d(f,g')tag5$$
The key points here are $(1)$ and the fact that if $A$ is a linear function then $partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.
answered 1 hour ago
Masacroso
11.9k41745
A slightly different approach.
Let $fin C^1(Bbb R^ntimes Bbb R^m,Bbb R^ell)$, then it is easy to check that
$$partial f(x,y)(a,b)=D_x f(x,y)a+D_y f(x,y)btag1$$
where $partial f$ is the Fréchet derivative of $f$ and $D_x f$ is the Fréchet derivative of $f(cdot,y)$. Similarly $D_y f$ is the Fréchet derivative of $f(x,cdot)$.
Let $d:Bbb R^ntimes Bbb R^ntoBbb R$ a dot product, hence
$$partial, d(x,y)(a,b)=D_x d(x,y)a+D_y d(x,y)btag2$$
Now note that the functions $d(cdot ,y)$ and $d(x,cdot)$ are linear, so
$$D_x d(x,y)=d(cdot, y)implies D_x d(x,y)a=d(a,y)tag3$$
And similarly $D_y d(x,y)b=d(x,b)$. Putting all together we find that
$$partial, d(x,y)(a,b)=d(a,y)+d(x,b)tag4$$
By last, using the chain rule, we have that
$$partial, [d(f,g)]=partial d(f,g)partial(f,g)=partial d(f,g)(f',g')\=d(f',g)+d(f,g')tag5$$
The key points here are $(1)$ and the fact that if $A$ is a linear function then $partial Ax=A$, what can be checked easily using the definition of Fréchet derivative.
answered 1 hour ago
Masacroso
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edited 18 mins ago
answered 1 hour ago
Masacroso
11.9k41745
answered 1 hour ago
Masacroso
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answered 1 hour ago
Masacroso
11.9k41745
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Do you know the definition of differentiability of a function $F:Bbb R^mtoBbb R^n$?
– edm
1 hour ago
1
What is the meaning of $<ldots>$?
– manooooh
1 hour ago
@edm A function is differentiable at t if there are functions $f_1, f_2, ..., f_n$ such that $f'(t) = (f_1'(t),...,f_n'(t))$ Can't really see how to apply that though :)
– Milan Leonard
1 hour ago