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Partial derivative - problem with understanding
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sorry for such basic question but I have problem with understanding partial derivative.
I watched some videos about that, and read some articles, bo with no result.
As I understand partial derivative gives me functions which are tengent to my main (parent) function.
In all tutorials that I've seen, they give example for $f(x,y)$. What you can write as $z = f(x,y)$
But I need tangent line to function of just one variable, like $f(x)$
So it seems to be obvious to just set $z=0$
So let's say my function is $f(x) = x^2$
I can write it as $y = x^2$
So after partial derivative on $x$, I get:
$y=2x$
And it's definetaly not tangent to $y=x^2$
I see I make some logical error, but don't know where and how to fix it.
Could anyone help me please?
Great thanks in advance
calculus multivariable-calculus partial-derivative
edited Jul 16 at 9:47
Nosrati
19.8k41644
asked Jul 16 at 9:44
pajczur
252
add a comment |Â
up vote
0
down vote
favorite
sorry for such basic question but I have problem with understanding partial derivative.
I watched some videos about that, and read some articles, bo with no result.
As I understand partial derivative gives me functions which are tengent to my main (parent) function.
In all tutorials that I've seen, they give example for $f(x,y)$. What you can write as $z = f(x,y)$
But I need tangent line to function of just one variable, like $f(x)$
So it seems to be obvious to just set $z=0$
So let's say my function is $f(x) = x^2$
I can write it as $y = x^2$
So after partial derivative on $x$, I get:
$y=2x$
And it's definetaly not tangent to $y=x^2$
I see I make some logical error, but don't know where and how to fix it.
Could anyone help me please?
Great thanks in advance
calculus multivariable-calculus partial-derivative
edited Jul 16 at 9:47
Nosrati
19.8k41644
asked Jul 16 at 9:44
pajczur
252
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
sorry for such basic question but I have problem with understanding partial derivative.
I watched some videos about that, and read some articles, bo with no result.
As I understand partial derivative gives me functions which are tengent to my main (parent) function.
In all tutorials that I've seen, they give example for $f(x,y)$. What you can write as $z = f(x,y)$
But I need tangent line to function of just one variable, like $f(x)$
So it seems to be obvious to just set $z=0$
So let's say my function is $f(x) = x^2$
I can write it as $y = x^2$
So after partial derivative on $x$, I get:
$y=2x$
And it's definetaly not tangent to $y=x^2$
I see I make some logical error, but don't know where and how to fix it.
Could anyone help me please?
Great thanks in advance
calculus multivariable-calculus partial-derivative
edited Jul 16 at 9:47
Nosrati
19.8k41644
asked Jul 16 at 9:44
pajczur
252
sorry for such basic question but I have problem with understanding partial derivative.
I watched some videos about that, and read some articles, bo with no result.
As I understand partial derivative gives me functions which are tengent to my main (parent) function.
In all tutorials that I've seen, they give example for $f(x,y)$. What you can write as $z = f(x,y)$
But I need tangent line to function of just one variable, like $f(x)$
So it seems to be obvious to just set $z=0$
So let's say my function is $f(x) = x^2$
I can write it as $y = x^2$
So after partial derivative on $x$, I get:
$y=2x$
And it's definetaly not tangent to $y=x^2$
I see I make some logical error, but don't know where and how to fix it.
Could anyone help me please?
Great thanks in advance
calculus multivariable-calculus partial-derivative
edited Jul 16 at 9:47
Nosrati
19.8k41644
asked Jul 16 at 9:44
pajczur
252
edited Jul 16 at 9:47
Nosrati
19.8k41644
edited Jul 16 at 9:47
Nosrati
19.8k41644
edited Jul 16 at 9:47
Nosrati
19.8k41644
19.8k41644
asked Jul 16 at 9:44
pajczur
252
asked Jul 16 at 9:44
pajczur
252
asked Jul 16 at 9:44
pajczur
252
252
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable:
$$f(x)=x^2$$
The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6left(x-3right) iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.
answered Jul 16 at 9:52
StackTD
20.1k1443
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
add a comment |Â
up vote
1
down vote
Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.
Concerning your specific example ($y=x^2$), the derivative is $2x$, yes. What that means is that the slope of the tangent line os the graph of $y=x^2$ that passes through $(x_0,x_0^2)$ is $2x_0$. And you can check that, indeed, the line $y=2x_0(x-x_0)+x_0^2(=2x_0x-x_0^2)$ is indeed tangente to the graph of that function. See the image below.
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable:
$$f(x)=x^2$$
The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6left(x-3right) iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.
answered Jul 16 at 9:52
StackTD
20.1k1443
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
add a comment |Â
up vote
0
down vote
accepted
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable:
$$f(x)=x^2$$
The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6left(x-3right) iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.
answered Jul 16 at 9:52
StackTD
20.1k1443
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable:
$$f(x)=x^2$$
The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6left(x-3right) iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.
answered Jul 16 at 9:52
StackTD
20.1k1443
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable:
$$f(x)=x^2$$
The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6left(x-3right) iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.
answered Jul 16 at 9:52
StackTD
20.1k1443
edited Jul 16 at 10:06
answered Jul 16 at 9:52
StackTD
20.1k1443
answered Jul 16 at 9:52
StackTD
20.1k1443
answered Jul 16 at 9:52
StackTD
20.1k1443
20.1k1443
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
add a comment |Â
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them.
– pajczur
Jul 16 at 10:21
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative.
– StackTD
Jul 16 at 11:23
add a comment |Â
up vote
1
down vote
Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.
Concerning your specific example ($y=x^2$), the derivative is $2x$, yes. What that means is that the slope of the tangent line os the graph of $y=x^2$ that passes through $(x_0,x_0^2)$ is $2x_0$. And you can check that, indeed, the line $y=2x_0(x-x_0)+x_0^2(=2x_0x-x_0^2)$ is indeed tangente to the graph of that function. See the image below.
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
add a comment |Â
up vote
1
down vote
Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.
Concerning your specific example ($y=x^2$), the derivative is $2x$, yes. What that means is that the slope of the tangent line os the graph of $y=x^2$ that passes through $(x_0,x_0^2)$ is $2x_0$. And you can check that, indeed, the line $y=2x_0(x-x_0)+x_0^2(=2x_0x-x_0^2)$ is indeed tangente to the graph of that function. See the image below.
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.
Concerning your specific example ($y=x^2$), the derivative is $2x$, yes. What that means is that the slope of the tangent line os the graph of $y=x^2$ that passes through $(x_0,x_0^2)$ is $2x_0$. And you can check that, indeed, the line $y=2x_0(x-x_0)+x_0^2(=2x_0x-x_0^2)$ is indeed tangente to the graph of that function. See the image below.
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.
Concerning your specific example ($y=x^2$), the derivative is $2x$, yes. What that means is that the slope of the tangent line os the graph of $y=x^2$ that passes through $(x_0,x_0^2)$ is $2x_0$. And you can check that, indeed, the line $y=2x_0(x-x_0)+x_0^2(=2x_0x-x_0^2)$ is indeed tangente to the graph of that function. See the image below.
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
edited Jul 16 at 10:32
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
answered Jul 16 at 9:49
José Carlos Santos
114k1698177
114k1698177
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
add a comment |Â
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula?
– pajczur
Jul 16 at 10:49
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b=x_0^2$, and $m=2x_0$.
– José Carlos Santos
Jul 16 at 10:52
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
:) now it seems to be obvious, thanks for your patience.
– pajczur
Jul 16 at 12:10
add a comment |Â
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