Differentiating a function that is defined generally (for eg, $ f (x_1, x_2, … x_n) $)
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If I'm asked to differentiate a function that is simply of the form: $ f(tx_1, tx_2, ... tx_n) $ (i.e., the function is not "defined" to be particularly anything), how should I go about this? I want to differentiate with respect to $ t$.
Context: Let's say we have a homogeneous function of with degree $ k $.
Thus,
$$ f( tx_1, tx_2, ... tx_n) = t^k f(x_1, x_2, ... x_n) $$
Differentiating on both sides with respect to $ t$, we get,
$$ x_1 f_1 (tx_1, tx_2, ... tx_n) + ... + x_n f_n (tx_1, tx_2, ... tx_n) = k cdot t^{k-1} cdot f(x_1, .. x_n) $$
On the RHS, why we differnetiate with respect to to get $ k cdot t^{k-1} $. Because the function on the RHS is a constant (i.e., it doesn't depend on t, we write it as it is)
On the LHS, why do we have: $ x_n f_n(tx_1, tx_2,dots, tx_n)$ for each term?
multivariable-calculus derivatives implicit-differentiation
edited Jan 17 at 11:33
Christoph
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
$endgroup$
add a comment |
$begingroup$
If I'm asked to differentiate a function that is simply of the form: $ f(tx_1, tx_2, ... tx_n) $ (i.e., the function is not "defined" to be particularly anything), how should I go about this? I want to differentiate with respect to $ t$.
Context: Let's say we have a homogeneous function of with degree $ k $.
Thus,
$$ f( tx_1, tx_2, ... tx_n) = t^k f(x_1, x_2, ... x_n) $$
Differentiating on both sides with respect to $ t$, we get,
$$ x_1 f_1 (tx_1, tx_2, ... tx_n) + ... + x_n f_n (tx_1, tx_2, ... tx_n) = k cdot t^{k-1} cdot f(x_1, .. x_n) $$
On the RHS, why we differnetiate with respect to to get $ k cdot t^{k-1} $. Because the function on the RHS is a constant (i.e., it doesn't depend on t, we write it as it is)
On the LHS, why do we have: $ x_n f_n(tx_1, tx_2,dots, tx_n)$ for each term?
multivariable-calculus derivatives implicit-differentiation
edited Jan 17 at 11:33
Christoph
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
$endgroup$
$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27
add a comment |
$begingroup$
If I'm asked to differentiate a function that is simply of the form: $ f(tx_1, tx_2, ... tx_n) $ (i.e., the function is not "defined" to be particularly anything), how should I go about this? I want to differentiate with respect to $ t$.
Context: Let's say we have a homogeneous function of with degree $ k $.
Thus,
$$ f( tx_1, tx_2, ... tx_n) = t^k f(x_1, x_2, ... x_n) $$
Differentiating on both sides with respect to $ t$, we get,
$$ x_1 f_1 (tx_1, tx_2, ... tx_n) + ... + x_n f_n (tx_1, tx_2, ... tx_n) = k cdot t^{k-1} cdot f(x_1, .. x_n) $$
On the RHS, why we differnetiate with respect to to get $ k cdot t^{k-1} $. Because the function on the RHS is a constant (i.e., it doesn't depend on t, we write it as it is)
On the LHS, why do we have: $ x_n f_n(tx_1, tx_2,dots, tx_n)$ for each term?
multivariable-calculus derivatives implicit-differentiation
edited Jan 17 at 11:33
Christoph
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
$endgroup$
If I'm asked to differentiate a function that is simply of the form: $ f(tx_1, tx_2, ... tx_n) $ (i.e., the function is not "defined" to be particularly anything), how should I go about this? I want to differentiate with respect to $ t$.
Context: Let's say we have a homogeneous function of with degree $ k $.
Thus,
$$ f( tx_1, tx_2, ... tx_n) = t^k f(x_1, x_2, ... x_n) $$
Differentiating on both sides with respect to $ t$, we get,
$$ x_1 f_1 (tx_1, tx_2, ... tx_n) + ... + x_n f_n (tx_1, tx_2, ... tx_n) = k cdot t^{k-1} cdot f(x_1, .. x_n) $$
On the RHS, why we differnetiate with respect to to get $ k cdot t^{k-1} $. Because the function on the RHS is a constant (i.e., it doesn't depend on t, we write it as it is)
On the LHS, why do we have: $ x_n f_n(tx_1, tx_2,dots, tx_n)$ for each term?
multivariable-calculus derivatives implicit-differentiation
multivariable-calculus derivatives implicit-differentiation
edited Jan 17 at 11:33
Christoph
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
edited Jan 17 at 11:33
Christoph
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
edited Jan 17 at 11:33
Christoph
12k1642
edited Jan 17 at 11:33
Christoph
12k1642
edited Jan 17 at 11:33
Christoph
12k1642
12k1642
asked Jan 17 at 11:19
WorldGovWorldGov
305111
asked Jan 17 at 11:19
WorldGovWorldGov
305111
asked Jan 17 at 11:19
WorldGovWorldGov
305111
305111
$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27
add a comment |
$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27
$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27
add a comment |
1 Answer
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$begingroup$
On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,dots,x_n)$ is constant with respect to $t$, so you can just keep it.
For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),dots,z_n(t)$ each of one variable $t$, then
$$
frac{mathrm d}{mathrm dt} f(z_1(t),dots,z_n(t)) = frac{partial f}{partial x_1}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_1}{mathrm dt}(t) + dots +frac{partial f}{partial x_n}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_n}{mathrm dt}(t)
$$
In your example you have $z_i(t) = tx_i$ so that $frac{mathrm dz_i}{mathrm dt}(t) = x_i$.
answered Jan 17 at 11:32
ChristophChristoph
12k1642
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add a comment |
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1 Answer
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$begingroup$
On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,dots,x_n)$ is constant with respect to $t$, so you can just keep it.
For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),dots,z_n(t)$ each of one variable $t$, then
$$
frac{mathrm d}{mathrm dt} f(z_1(t),dots,z_n(t)) = frac{partial f}{partial x_1}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_1}{mathrm dt}(t) + dots +frac{partial f}{partial x_n}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_n}{mathrm dt}(t)
$$
In your example you have $z_i(t) = tx_i$ so that $frac{mathrm dz_i}{mathrm dt}(t) = x_i$.
answered Jan 17 at 11:32
ChristophChristoph
12k1642
$endgroup$
add a comment |
$begingroup$
On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,dots,x_n)$ is constant with respect to $t$, so you can just keep it.
For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),dots,z_n(t)$ each of one variable $t$, then
$$
frac{mathrm d}{mathrm dt} f(z_1(t),dots,z_n(t)) = frac{partial f}{partial x_1}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_1}{mathrm dt}(t) + dots +frac{partial f}{partial x_n}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_n}{mathrm dt}(t)
$$
In your example you have $z_i(t) = tx_i$ so that $frac{mathrm dz_i}{mathrm dt}(t) = x_i$.
answered Jan 17 at 11:32
ChristophChristoph
12k1642
$endgroup$
add a comment |
$begingroup$
On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,dots,x_n)$ is constant with respect to $t$, so you can just keep it.
For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),dots,z_n(t)$ each of one variable $t$, then
$$
frac{mathrm d}{mathrm dt} f(z_1(t),dots,z_n(t)) = frac{partial f}{partial x_1}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_1}{mathrm dt}(t) + dots +frac{partial f}{partial x_n}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_n}{mathrm dt}(t)
$$
In your example you have $z_i(t) = tx_i$ so that $frac{mathrm dz_i}{mathrm dt}(t) = x_i$.
answered Jan 17 at 11:32
ChristophChristoph
12k1642
$endgroup$
On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,dots,x_n)$ is constant with respect to $t$, so you can just keep it.
For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),dots,z_n(t)$ each of one variable $t$, then
$$
frac{mathrm d}{mathrm dt} f(z_1(t),dots,z_n(t)) = frac{partial f}{partial x_1}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_1}{mathrm dt}(t) + dots +frac{partial f}{partial x_n}(z_1(t),dots,z_n(t)) cdot frac{mathrm dz_n}{mathrm dt}(t)
$$
In your example you have $z_i(t) = tx_i$ so that $frac{mathrm dz_i}{mathrm dt}(t) = x_i$.
answered Jan 17 at 11:32
ChristophChristoph
12k1642
edited Jan 17 at 11:38
answered Jan 17 at 11:32
ChristophChristoph
12k1642
answered Jan 17 at 11:32
ChristophChristoph
12k1642
answered Jan 17 at 11:32
ChristophChristoph
12k1642
12k1642
add a comment |
add a comment |
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$begingroup$
Shouldn't your RHS be $k cdot t^{k-1} cdot f(x_1, dots, x_n)$? (without the $t$ inside)
$endgroup$
– BigbearZzz
Jan 17 at 11:22
$begingroup$
Yes, sorry. That was a mistake on my part.
$endgroup$
– WorldGov
Jan 17 at 11:23
$begingroup$
You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple.
$endgroup$
– James
Jan 17 at 11:27