Mixing
Question about the chain rule.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I was doing some practice problems with implicit differentiation and the chain rule. This is something that I have known how to do, but I do not understand why it works. Here is a screenshot of the derivative of $y^2$ with respect to $x$:
Where exactly does the $fracdydxfracddy$ come from? Why exactly does $dy$ stay in the numerator; why is the final term $fracdydx[2y]$ and not $fracddx[2x]$?
calculus derivatives implicit-differentiation
asked 1 hour ago
nine-hundred
1147
add a comment |Â
up vote
1
down vote
favorite
I was doing some practice problems with implicit differentiation and the chain rule. This is something that I have known how to do, but I do not understand why it works. Here is a screenshot of the derivative of $y^2$ with respect to $x$:
Where exactly does the $fracdydxfracddy$ come from? Why exactly does $dy$ stay in the numerator; why is the final term $fracdydx[2y]$ and not $fracddx[2x]$?
calculus derivatives implicit-differentiation
asked 1 hour ago
nine-hundred
1147
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was doing some practice problems with implicit differentiation and the chain rule. This is something that I have known how to do, but I do not understand why it works. Here is a screenshot of the derivative of $y^2$ with respect to $x$:
Where exactly does the $fracdydxfracddy$ come from? Why exactly does $dy$ stay in the numerator; why is the final term $fracdydx[2y]$ and not $fracddx[2x]$?
calculus derivatives implicit-differentiation
asked 1 hour ago
nine-hundred
1147
I was doing some practice problems with implicit differentiation and the chain rule. This is something that I have known how to do, but I do not understand why it works. Here is a screenshot of the derivative of $y^2$ with respect to $x$:
Where exactly does the $fracdydxfracddy$ come from? Why exactly does $dy$ stay in the numerator; why is the final term $fracdydx[2y]$ and not $fracddx[2x]$?
calculus derivatives implicit-differentiation
calculus derivatives implicit-differentiation
asked 1 hour ago
nine-hundred
1147
asked 1 hour ago
nine-hundred
1147
asked 1 hour ago
nine-hundred
1147
asked 1 hour ago
nine-hundred
1147
asked 1 hour ago
nine-hundred
1147
1147
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
It might be easier to unerstand with the "prime" notation instead of Leibniz notation.
The chain rule says that $f'(g(x)) = g'(x) f'(g(x))$. In your case, $y(x) = g(x)$ and $f(x) = x^2$. You want to find the derivative of $f(g(x))$, which by the chain rule is $g'(x) f'(g(x))$, or $y'(x) cdot 2g(x)$
answered 48 mins ago
Ovi
11.6k936105
add a comment |Â
up vote
2
down vote
If you want to see why this works fundamentally, we can derive it from the definition of derivatives.
$$
lim_hto0 fracf(x+h)^2 - f(x)^2h = lim_hto0 frac(f(x+h) - f(x))h(f(x+h)+f(x))
$$
The first term becomes $f'(x)$ and the second term becomes $2f(x)$. So the limit is $2f'(x)f(x)$. If $y=f(x)$, you can convert this to Leibniz's notation to get $fracdydx[2y]$.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
The term $fracdydxfracddy$ comes from the chain rule. Since you want to differentiate with respect to $x$ but your function is interms of $y$, by the chain rule first you differentiate your function with respect to $y$ and then multiply by the derivative of $y$ with respect to x. The notation $fracddx$ is kind of nice since it can be seen like you are doing some kind of multiplication by $1$
$$fracdy^2dx=fracdy^2dxcdot 1=fracdy^2dxfracdydy=fracdy^2dyfracdydx=2yfracdydx.$$
Though I treated $fracddx$ as fractions ,$fracddx$ is not really a fraction but turns out it works like a fraction when you apply the chain rule.
answered 50 mins ago
Basanta R Pahari
1,450513
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
It might be easier to unerstand with the "prime" notation instead of Leibniz notation.
The chain rule says that $f'(g(x)) = g'(x) f'(g(x))$. In your case, $y(x) = g(x)$ and $f(x) = x^2$. You want to find the derivative of $f(g(x))$, which by the chain rule is $g'(x) f'(g(x))$, or $y'(x) cdot 2g(x)$
answered 48 mins ago
Ovi
11.6k936105
add a comment |Â
up vote
2
down vote
accepted
It might be easier to unerstand with the "prime" notation instead of Leibniz notation.
The chain rule says that $f'(g(x)) = g'(x) f'(g(x))$. In your case, $y(x) = g(x)$ and $f(x) = x^2$. You want to find the derivative of $f(g(x))$, which by the chain rule is $g'(x) f'(g(x))$, or $y'(x) cdot 2g(x)$
answered 48 mins ago
Ovi
11.6k936105
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It might be easier to unerstand with the "prime" notation instead of Leibniz notation.
The chain rule says that $f'(g(x)) = g'(x) f'(g(x))$. In your case, $y(x) = g(x)$ and $f(x) = x^2$. You want to find the derivative of $f(g(x))$, which by the chain rule is $g'(x) f'(g(x))$, or $y'(x) cdot 2g(x)$
answered 48 mins ago
Ovi
11.6k936105
It might be easier to unerstand with the "prime" notation instead of Leibniz notation.
The chain rule says that $f'(g(x)) = g'(x) f'(g(x))$. In your case, $y(x) = g(x)$ and $f(x) = x^2$. You want to find the derivative of $f(g(x))$, which by the chain rule is $g'(x) f'(g(x))$, or $y'(x) cdot 2g(x)$
answered 48 mins ago
Ovi
11.6k936105
edited 29 mins ago
answered 48 mins ago
Ovi
11.6k936105
answered 48 mins ago
Ovi
11.6k936105
answered 48 mins ago
Ovi
11.6k936105
11.6k936105
add a comment |Â
add a comment |Â
up vote
2
down vote
If you want to see why this works fundamentally, we can derive it from the definition of derivatives.
$$
lim_hto0 fracf(x+h)^2 - f(x)^2h = lim_hto0 frac(f(x+h) - f(x))h(f(x+h)+f(x))
$$
The first term becomes $f'(x)$ and the second term becomes $2f(x)$. So the limit is $2f'(x)f(x)$. If $y=f(x)$, you can convert this to Leibniz's notation to get $fracdydx[2y]$.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
If you want to see why this works fundamentally, we can derive it from the definition of derivatives.
$$
lim_hto0 fracf(x+h)^2 - f(x)^2h = lim_hto0 frac(f(x+h) - f(x))h(f(x+h)+f(x))
$$
The first term becomes $f'(x)$ and the second term becomes $2f(x)$. So the limit is $2f'(x)f(x)$. If $y=f(x)$, you can convert this to Leibniz's notation to get $fracdydx[2y]$.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you want to see why this works fundamentally, we can derive it from the definition of derivatives.
$$
lim_hto0 fracf(x+h)^2 - f(x)^2h = lim_hto0 frac(f(x+h) - f(x))h(f(x+h)+f(x))
$$
The first term becomes $f'(x)$ and the second term becomes $2f(x)$. So the limit is $2f'(x)f(x)$. If $y=f(x)$, you can convert this to Leibniz's notation to get $fracdydx[2y]$.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you want to see why this works fundamentally, we can derive it from the definition of derivatives.
$$
lim_hto0 fracf(x+h)^2 - f(x)^2h = lim_hto0 frac(f(x+h) - f(x))h(f(x+h)+f(x))
$$
The first term becomes $f'(x)$ and the second term becomes $2f(x)$. So the limit is $2f'(x)f(x)$. If $y=f(x)$, you can convert this to Leibniz's notation to get $fracdydx[2y]$.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 44 mins ago
noocsharp
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 44 mins ago
noocsharp
212
answered 44 mins ago
noocsharp
212
212
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
noocsharp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
1
down vote
The term $fracdydxfracddy$ comes from the chain rule. Since you want to differentiate with respect to $x$ but your function is interms of $y$, by the chain rule first you differentiate your function with respect to $y$ and then multiply by the derivative of $y$ with respect to x. The notation $fracddx$ is kind of nice since it can be seen like you are doing some kind of multiplication by $1$
$$fracdy^2dx=fracdy^2dxcdot 1=fracdy^2dxfracdydy=fracdy^2dyfracdydx=2yfracdydx.$$
Though I treated $fracddx$ as fractions ,$fracddx$ is not really a fraction but turns out it works like a fraction when you apply the chain rule.
answered 50 mins ago
Basanta R Pahari
1,450513
add a comment |Â
up vote
1
down vote
The term $fracdydxfracddy$ comes from the chain rule. Since you want to differentiate with respect to $x$ but your function is interms of $y$, by the chain rule first you differentiate your function with respect to $y$ and then multiply by the derivative of $y$ with respect to x. The notation $fracddx$ is kind of nice since it can be seen like you are doing some kind of multiplication by $1$
$$fracdy^2dx=fracdy^2dxcdot 1=fracdy^2dxfracdydy=fracdy^2dyfracdydx=2yfracdydx.$$
Though I treated $fracddx$ as fractions ,$fracddx$ is not really a fraction but turns out it works like a fraction when you apply the chain rule.
answered 50 mins ago
Basanta R Pahari
1,450513
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The term $fracdydxfracddy$ comes from the chain rule. Since you want to differentiate with respect to $x$ but your function is interms of $y$, by the chain rule first you differentiate your function with respect to $y$ and then multiply by the derivative of $y$ with respect to x. The notation $fracddx$ is kind of nice since it can be seen like you are doing some kind of multiplication by $1$
$$fracdy^2dx=fracdy^2dxcdot 1=fracdy^2dxfracdydy=fracdy^2dyfracdydx=2yfracdydx.$$
Though I treated $fracddx$ as fractions ,$fracddx$ is not really a fraction but turns out it works like a fraction when you apply the chain rule.
answered 50 mins ago
Basanta R Pahari
1,450513
The term $fracdydxfracddy$ comes from the chain rule. Since you want to differentiate with respect to $x$ but your function is interms of $y$, by the chain rule first you differentiate your function with respect to $y$ and then multiply by the derivative of $y$ with respect to x. The notation $fracddx$ is kind of nice since it can be seen like you are doing some kind of multiplication by $1$
$$fracdy^2dx=fracdy^2dxcdot 1=fracdy^2dxfracdydy=fracdy^2dyfracdydx=2yfracdydx.$$
Though I treated $fracddx$ as fractions ,$fracddx$ is not really a fraction but turns out it works like a fraction when you apply the chain rule.
answered 50 mins ago
Basanta R Pahari
1,450513
answered 50 mins ago
Basanta R Pahari
1,450513
answered 50 mins ago
Basanta R Pahari
1,450513
answered 50 mins ago
Basanta R Pahari
1,450513
1,450513
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2940216%2fquestion-about-the-chain-rule%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password