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A proof of the product rule using the single variable chain rule?
Clash Royale CLAN TAG#URR8PPP
up vote
12
down vote
favorite
A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?
calculus alternative-proof
asked Aug 2 at 19:27
Aaron
15.1k22552
add a comment |Â
up vote
12
down vote
favorite
A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?
calculus alternative-proof
asked Aug 2 at 19:27
Aaron
15.1k22552
Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
2
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?
calculus alternative-proof
asked Aug 2 at 19:27
Aaron
15.1k22552
A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?
calculus alternative-proof
asked Aug 2 at 19:27
Aaron
15.1k22552
asked Aug 2 at 19:27
Aaron
15.1k22552
asked Aug 2 at 19:27
Aaron
15.1k22552
asked Aug 2 at 19:27
Aaron
15.1k22552
15.1k22552
Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
2
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36
add a comment |Â
Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
2
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36
Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
2
2
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
32
down vote
accepted
Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.
Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule,
$$f^prime = 2(u+v)(u^prime+v^prime)=2(uu^prime+uv^prime+vu^prime+vv^prime)text.$$
Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have
$$f^prime=2uu^prime+2(uv)^prime+2vv^prime=2[uu^prime+(uv)^prime+vv^prime]text.$$
It follows immediately that
$$(uv)^prime=uv^prime+vu^primetext.$$
answered Aug 2 at 19:58
Clarinetist
10.3k32767
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
add a comment |Â
up vote
7
down vote
Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.
Consider two functions $f,g$ and the expression $fracddx (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).
Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $fracddx(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.
Since these two expressions are equal, we have
$$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$
Using some algebra,
$$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$
$$2fg’+2f’g=2(fg)’$$
$$(fg)’=fg’+f’g$$
answered Aug 2 at 20:06
Tyler6
440210
add a comment |Â
up vote
6
down vote
It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $log(fg) = log(f) + log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
32
down vote
accepted
Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.
Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule,
$$f^prime = 2(u+v)(u^prime+v^prime)=2(uu^prime+uv^prime+vu^prime+vv^prime)text.$$
Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have
$$f^prime=2uu^prime+2(uv)^prime+2vv^prime=2[uu^prime+(uv)^prime+vv^prime]text.$$
It follows immediately that
$$(uv)^prime=uv^prime+vu^primetext.$$
answered Aug 2 at 19:58
Clarinetist
10.3k32767
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
add a comment |Â
up vote
32
down vote
accepted
Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.
Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule,
$$f^prime = 2(u+v)(u^prime+v^prime)=2(uu^prime+uv^prime+vu^prime+vv^prime)text.$$
Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have
$$f^prime=2uu^prime+2(uv)^prime+2vv^prime=2[uu^prime+(uv)^prime+vv^prime]text.$$
It follows immediately that
$$(uv)^prime=uv^prime+vu^primetext.$$
answered Aug 2 at 19:58
Clarinetist
10.3k32767
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
add a comment |Â
up vote
32
down vote
accepted
up vote
32
down vote
accepted
Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.
Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule,
$$f^prime = 2(u+v)(u^prime+v^prime)=2(uu^prime+uv^prime+vu^prime+vv^prime)text.$$
Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have
$$f^prime=2uu^prime+2(uv)^prime+2vv^prime=2[uu^prime+(uv)^prime+vv^prime]text.$$
It follows immediately that
$$(uv)^prime=uv^prime+vu^primetext.$$
answered Aug 2 at 19:58
Clarinetist
10.3k32767
Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.
Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule,
$$f^prime = 2(u+v)(u^prime+v^prime)=2(uu^prime+uv^prime+vu^prime+vv^prime)text.$$
Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have
$$f^prime=2uu^prime+2(uv)^prime+2vv^prime=2[uu^prime+(uv)^prime+vv^prime]text.$$
It follows immediately that
$$(uv)^prime=uv^prime+vu^primetext.$$
answered Aug 2 at 19:58
Clarinetist
10.3k32767
edited Aug 2 at 20:01
answered Aug 2 at 19:58
Clarinetist
10.3k32767
answered Aug 2 at 19:58
Clarinetist
10.3k32767
answered Aug 2 at 19:58
Clarinetist
10.3k32767
10.3k32767
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
add a comment |Â
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
2
2
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
Very nice. Exactly the kind of thing I was hoping for (since it only relies on the derivative for $x^2$ and the chain rule.
– Aaron
Aug 2 at 20:01
add a comment |Â
up vote
7
down vote
Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.
Consider two functions $f,g$ and the expression $fracddx (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).
Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $fracddx(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.
Since these two expressions are equal, we have
$$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$
Using some algebra,
$$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$
$$2fg’+2f’g=2(fg)’$$
$$(fg)’=fg’+f’g$$
answered Aug 2 at 20:06
Tyler6
440210
add a comment |Â
up vote
7
down vote
Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.
Consider two functions $f,g$ and the expression $fracddx (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).
Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $fracddx(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.
Since these two expressions are equal, we have
$$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$
Using some algebra,
$$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$
$$2fg’+2f’g=2(fg)’$$
$$(fg)’=fg’+f’g$$
answered Aug 2 at 20:06
Tyler6
440210
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.
Consider two functions $f,g$ and the expression $fracddx (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).
Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $fracddx(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.
Since these two expressions are equal, we have
$$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$
Using some algebra,
$$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$
$$2fg’+2f’g=2(fg)’$$
$$(fg)’=fg’+f’g$$
answered Aug 2 at 20:06
Tyler6
440210
Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.
Consider two functions $f,g$ and the expression $fracddx (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).
Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $fracddx(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.
Since these two expressions are equal, we have
$$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$
Using some algebra,
$$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$
$$2fg’+2f’g=2(fg)’$$
$$(fg)’=fg’+f’g$$
answered Aug 2 at 20:06
Tyler6
440210
answered Aug 2 at 20:06
Tyler6
440210
answered Aug 2 at 20:06
Tyler6
440210
answered Aug 2 at 20:06
Tyler6
440210
440210
add a comment |Â
add a comment |Â
up vote
6
down vote
It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $log(fg) = log(f) + log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
 |Â
show 5 more comments
up vote
6
down vote
It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $log(fg) = log(f) + log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
 |Â
show 5 more comments
up vote
6
down vote
up vote
6
down vote
It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $log(fg) = log(f) + log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $log(fg) = log(f) + log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
answered Aug 2 at 19:31
Sridhar Ramesh
1,238715
1,238715
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
 |Â
show 5 more comments
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
2
2
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Yes, that is the easy proof using log differentiation that I said I already knew.
– Aaron
Aug 2 at 19:31
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
Ah, sorry, I read too fast and misinterpreted. But note that the proof using logarithmic differentiation is using the single-variable chain rule, in its differentiations of log(whatever).
– Sridhar Ramesh
Aug 2 at 19:35
1
1
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
That's perfectly fine. Log differentiation is technically an application of the chain rule, so you aren't wrong.
– Aaron
Aug 2 at 19:36
1
1
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
@Aaron: This is not perfectly fine. You cannot take logarithm of anything unless you know it is nonzero. Furthermore, it is false that $ln(xy) = ln(x)+ln(y)$ when $x=y=-1$, for any reasonable definition of $ln$.
– user21820
Aug 3 at 8:46
1
1
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
@user21820 I've taught calculus enough times to know exactly what goes into the proofs here. You don't need the product rule (though I think you need linearity of the derivative). Yes, circularity can be subtle. It just doesn't happen here. You don't need the product rule to establish the definition of integration, the fundamental theorem of calculus, u-substitution, the definition of log as an integral, or the relevant property of logs. If you don't believe me, sit down and do it. Your doubt is meaningless in the face of something this easy to verify.
– Aaron
Aug 3 at 15:15
 |Â
show 5 more comments
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Just out of curiosity, is this where you saw the claim?
– User
Aug 3 at 7:43
@user Yes, although when I had last looked at that thread, nobody had responded to the comment.
– Aaron
Aug 3 at 12:13
2
related: math.stackexchange.com/a/208014/13618 mathoverflow.net/questions/44774/… The product rule and chain rule are not logically independent. Given one, you can prove the other/
– Ben Crowell
Aug 3 at 15:36