Confusion about implicit differentiation $frac{dy}{dx}$
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Today I learnt about implicit differentiation using this:
$frac{d}{dx}(f)$ = $frac{df}{dy} times frac{dy}{dx}$
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
$frac{d(y^2)}{dx} = frac{d(y^2)}{dy}timesfrac{dy}{dx}=2yfrac{dy}{dx}$
Why is the derivative of $y^2$ with respect to $y$, $2y$? Do you assume it's a function of something else?
And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:
$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$
How does that simplify to the $2x$ I know it is?
Thanks
differential-equations implicit-differentiation
add a comment |
up vote
2
down vote
favorite
Today I learnt about implicit differentiation using this:
$frac{d}{dx}(f)$ = $frac{df}{dy} times frac{dy}{dx}$
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
$frac{d(y^2)}{dx} = frac{d(y^2)}{dy}timesfrac{dy}{dx}=2yfrac{dy}{dx}$
Why is the derivative of $y^2$ with respect to $y$, $2y$? Do you assume it's a function of something else?
And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:
$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$
How does that simplify to the $2x$ I know it is?
Thanks
differential-equations implicit-differentiation
What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Today I learnt about implicit differentiation using this:
$frac{d}{dx}(f)$ = $frac{df}{dy} times frac{dy}{dx}$
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
$frac{d(y^2)}{dx} = frac{d(y^2)}{dy}timesfrac{dy}{dx}=2yfrac{dy}{dx}$
Why is the derivative of $y^2$ with respect to $y$, $2y$? Do you assume it's a function of something else?
And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:
$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$
How does that simplify to the $2x$ I know it is?
Thanks
differential-equations implicit-differentiation
Today I learnt about implicit differentiation using this:
$frac{d}{dx}(f)$ = $frac{df}{dy} times frac{dy}{dx}$
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
$frac{d(y^2)}{dx} = frac{d(y^2)}{dy}timesfrac{dy}{dx}=2yfrac{dy}{dx}$
Why is the derivative of $y^2$ with respect to $y$, $2y$? Do you assume it's a function of something else?
And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:
$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$
How does that simplify to the $2x$ I know it is?
Thanks
differential-equations implicit-differentiation
differential-equations implicit-differentiation
edited 10 hours ago
WesleyGroupshaveFeelingsToo
992321
992321
asked 10 hours ago
DevinJC
936
936
What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago
add a comment |
What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago
What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Yes, here they assume that $y$ is a function of something else.
Whenever we apply implicit differentiation like with $frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
$$ frac{d y^2(x)}{d x} =2 y(x) cdotfrac{d y(x) }{dx}$$
Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $frac{ d y}{dx} =0$ and the equation just reads $0=0$.
For your second case where you state:
$$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$$
We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get:
$$ frac{dy}{dx} =frac{dy}{dy}timesfrac{d x^2 }{dx}= 1 cdot 2x $$
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
$$x+ x^2 + 3y^2 + y^3 = 6$$
Using implicit differentiation we could construct the tangent line for $(x,y)=(1,1)$
We would get:
$$1+ 2x + 6y frac{dy}{dx} + 3 y^2 frac{dy}{dx} = 0$$
Now we plug in the value at (1,1) to get:
$$1+ 2 + 6 frac{dy}{dx} + 3 frac{dy}{dx} = 0$$
Or after rearranging:
$$9 frac{dy}{dx} = -3 rightarrow frac{dy}{dx}= -frac{1}{3}$$
We get the tangent line $y= -frac{1}{3}(x-1) +1$
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
add a comment |
up vote
4
down vote
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
$$
f = y(x)^2, qquadfrac{df}{dx}=frac{d(y^2)}{dx},\
f=g(y(x)),qquad g(y)=y^2,\
frac{d}{dx}g(y(x))=frac{dg(y(x))}{dy}frac{dy(x)}{dx}=(2y(x))frac{dy}{dx}
$$
add a comment |
up vote
2
down vote
The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative:
$$lim_{h to 0} frac{(y+h)^2-y^2}{h} = 2y.$$
In the last expression you write
$$frac{d(x^2)}{dx} = frac{d(x^2)}{dy}frac{dy}{dx}$$
but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is
$$frac{d(f^2)}{dy}frac{dy}{dx} = 2f(y)f'(y)frac{dy}{dx} = 2xf'(y)frac{dy}{dx} = 2xfrac{dx}{dy}frac{dy}{dx} = 2xfrac{dx}{dx} = 2x.$$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, here they assume that $y$ is a function of something else.
Whenever we apply implicit differentiation like with $frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
$$ frac{d y^2(x)}{d x} =2 y(x) cdotfrac{d y(x) }{dx}$$
Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $frac{ d y}{dx} =0$ and the equation just reads $0=0$.
For your second case where you state:
$$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$$
We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get:
$$ frac{dy}{dx} =frac{dy}{dy}timesfrac{d x^2 }{dx}= 1 cdot 2x $$
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
$$x+ x^2 + 3y^2 + y^3 = 6$$
Using implicit differentiation we could construct the tangent line for $(x,y)=(1,1)$
We would get:
$$1+ 2x + 6y frac{dy}{dx} + 3 y^2 frac{dy}{dx} = 0$$
Now we plug in the value at (1,1) to get:
$$1+ 2 + 6 frac{dy}{dx} + 3 frac{dy}{dx} = 0$$
Or after rearranging:
$$9 frac{dy}{dx} = -3 rightarrow frac{dy}{dx}= -frac{1}{3}$$
We get the tangent line $y= -frac{1}{3}(x-1) +1$
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
add a comment |
up vote
2
down vote
accepted
Yes, here they assume that $y$ is a function of something else.
Whenever we apply implicit differentiation like with $frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
$$ frac{d y^2(x)}{d x} =2 y(x) cdotfrac{d y(x) }{dx}$$
Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $frac{ d y}{dx} =0$ and the equation just reads $0=0$.
For your second case where you state:
$$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$$
We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get:
$$ frac{dy}{dx} =frac{dy}{dy}timesfrac{d x^2 }{dx}= 1 cdot 2x $$
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
$$x+ x^2 + 3y^2 + y^3 = 6$$
Using implicit differentiation we could construct the tangent line for $(x,y)=(1,1)$
We would get:
$$1+ 2x + 6y frac{dy}{dx} + 3 y^2 frac{dy}{dx} = 0$$
Now we plug in the value at (1,1) to get:
$$1+ 2 + 6 frac{dy}{dx} + 3 frac{dy}{dx} = 0$$
Or after rearranging:
$$9 frac{dy}{dx} = -3 rightarrow frac{dy}{dx}= -frac{1}{3}$$
We get the tangent line $y= -frac{1}{3}(x-1) +1$
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, here they assume that $y$ is a function of something else.
Whenever we apply implicit differentiation like with $frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
$$ frac{d y^2(x)}{d x} =2 y(x) cdotfrac{d y(x) }{dx}$$
Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $frac{ d y}{dx} =0$ and the equation just reads $0=0$.
For your second case where you state:
$$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$$
We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get:
$$ frac{dy}{dx} =frac{dy}{dy}timesfrac{d x^2 }{dx}= 1 cdot 2x $$
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
$$x+ x^2 + 3y^2 + y^3 = 6$$
Using implicit differentiation we could construct the tangent line for $(x,y)=(1,1)$
We would get:
$$1+ 2x + 6y frac{dy}{dx} + 3 y^2 frac{dy}{dx} = 0$$
Now we plug in the value at (1,1) to get:
$$1+ 2 + 6 frac{dy}{dx} + 3 frac{dy}{dx} = 0$$
Or after rearranging:
$$9 frac{dy}{dx} = -3 rightarrow frac{dy}{dx}= -frac{1}{3}$$
We get the tangent line $y= -frac{1}{3}(x-1) +1$
Yes, here they assume that $y$ is a function of something else.
Whenever we apply implicit differentiation like with $frac{dy^2(x)}{dx}$, we sort of first forget it is a function of something else, say $x$, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
$$ frac{d y^2(x)}{d x} =2 y(x) cdotfrac{d y(x) }{dx}$$
Actually implict differentiation often boils down to just using the chain rule. Notice that if $y$ is not a function of $x$, we get that $frac{ d y}{dx} =0$ and the equation just reads $0=0$.
For your second case where you state:
$$frac{d(x^2)}{dx} =frac{d(x^2)}{dy}timesfrac{dy}{dx}$$
We could call $y=x^2$ because you really need to tell the reader what this new variable $y$ is in relation to the old variable $x$, else the derivative could be anything. We then get:
$$ frac{dy}{dx} =frac{dy}{dy}timesfrac{d x^2 }{dx}= 1 cdot 2x $$
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
$$x+ x^2 + 3y^2 + y^3 = 6$$
Using implicit differentiation we could construct the tangent line for $(x,y)=(1,1)$
We would get:
$$1+ 2x + 6y frac{dy}{dx} + 3 y^2 frac{dy}{dx} = 0$$
Now we plug in the value at (1,1) to get:
$$1+ 2 + 6 frac{dy}{dx} + 3 frac{dy}{dx} = 0$$
Or after rearranging:
$$9 frac{dy}{dx} = -3 rightarrow frac{dy}{dx}= -frac{1}{3}$$
We get the tangent line $y= -frac{1}{3}(x-1) +1$
edited 10 hours ago
community wiki
4 revs
WesleyGroupshaveFeelingsToo
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
add a comment |
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
Thanks a lot this really helped.
– DevinJC
10 hours ago
Thanks a lot this really helped.
– DevinJC
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
any time ;) we're all here to help
– WesleyGroupshaveFeelingsToo
10 hours ago
add a comment |
up vote
4
down vote
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
$$
f = y(x)^2, qquadfrac{df}{dx}=frac{d(y^2)}{dx},\
f=g(y(x)),qquad g(y)=y^2,\
frac{d}{dx}g(y(x))=frac{dg(y(x))}{dy}frac{dy(x)}{dx}=(2y(x))frac{dy}{dx}
$$
add a comment |
up vote
4
down vote
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
$$
f = y(x)^2, qquadfrac{df}{dx}=frac{d(y^2)}{dx},\
f=g(y(x)),qquad g(y)=y^2,\
frac{d}{dx}g(y(x))=frac{dg(y(x))}{dy}frac{dy(x)}{dx}=(2y(x))frac{dy}{dx}
$$
add a comment |
up vote
4
down vote
up vote
4
down vote
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
$$
f = y(x)^2, qquadfrac{df}{dx}=frac{d(y^2)}{dx},\
f=g(y(x)),qquad g(y)=y^2,\
frac{d}{dx}g(y(x))=frac{dg(y(x))}{dy}frac{dy(x)}{dx}=(2y(x))frac{dy}{dx}
$$
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
$$
f = y(x)^2, qquadfrac{df}{dx}=frac{d(y^2)}{dx},\
f=g(y(x)),qquad g(y)=y^2,\
frac{d}{dx}g(y(x))=frac{dg(y(x))}{dy}frac{dy(x)}{dx}=(2y(x))frac{dy}{dx}
$$
answered 10 hours ago
Vasily Mitch
99817
99817
add a comment |
add a comment |
up vote
2
down vote
The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative:
$$lim_{h to 0} frac{(y+h)^2-y^2}{h} = 2y.$$
In the last expression you write
$$frac{d(x^2)}{dx} = frac{d(x^2)}{dy}frac{dy}{dx}$$
but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is
$$frac{d(f^2)}{dy}frac{dy}{dx} = 2f(y)f'(y)frac{dy}{dx} = 2xf'(y)frac{dy}{dx} = 2xfrac{dx}{dy}frac{dy}{dx} = 2xfrac{dx}{dx} = 2x.$$
add a comment |
up vote
2
down vote
The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative:
$$lim_{h to 0} frac{(y+h)^2-y^2}{h} = 2y.$$
In the last expression you write
$$frac{d(x^2)}{dx} = frac{d(x^2)}{dy}frac{dy}{dx}$$
but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is
$$frac{d(f^2)}{dy}frac{dy}{dx} = 2f(y)f'(y)frac{dy}{dx} = 2xf'(y)frac{dy}{dx} = 2xfrac{dx}{dy}frac{dy}{dx} = 2xfrac{dx}{dx} = 2x.$$
add a comment |
up vote
2
down vote
up vote
2
down vote
The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative:
$$lim_{h to 0} frac{(y+h)^2-y^2}{h} = 2y.$$
In the last expression you write
$$frac{d(x^2)}{dx} = frac{d(x^2)}{dy}frac{dy}{dx}$$
but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is
$$frac{d(f^2)}{dy}frac{dy}{dx} = 2f(y)f'(y)frac{dy}{dx} = 2xf'(y)frac{dy}{dx} = 2xfrac{dx}{dy}frac{dy}{dx} = 2xfrac{dx}{dx} = 2x.$$
The derivative of $y^2$ with respect to $y$ is $2y$ because of the definition of derivative:
$$lim_{h to 0} frac{(y+h)^2-y^2}{h} = 2y.$$
In the last expression you write
$$frac{d(x^2)}{dx} = frac{d(x^2)}{dy}frac{dy}{dx}$$
but this is meaningful only when $x$ is a function of $y$. So assume $x = f(y)$. Then, by the first equation you mention in the question, the right hand side is
$$frac{d(f^2)}{dy}frac{dy}{dx} = 2f(y)f'(y)frac{dy}{dx} = 2xf'(y)frac{dy}{dx} = 2xfrac{dx}{dy}frac{dy}{dx} = 2xfrac{dx}{dx} = 2x.$$
answered 10 hours ago
Gibbs
4,6922726
4,6922726
add a comment |
add a comment |
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What is $y$ in your last differentiation? Take $y=x$ then it works perfectly.
– A.Γ.
10 hours ago
@A.Γ. Its just an example, I'm not really sure, I thought of it as $y=x^2$ and $y=f$and did it using the formula at the top.
– DevinJC
10 hours ago
If $y=x^2$ it still works perfectly.
– A.Γ.
10 hours ago