Mixing
Derivative of total revenue
$begingroup$
I have a maths related question and I was hoping that you could help me out. I'm trying to understand why, when differentiating the total revenue function, one is required to differentiate implicitly.
The total revenue formula is $R=PQ$, where $R$ is total revenue, $P$ is the unit price and $Q$ is the quantity. I know how to differentiate implicitly... $frac{dR}{dP}=Q+Pfrac{dQ}{dP}$ ... but I don't quite understand why it's needed. That is, why don't we treat $Q$ like a constant, giving $frac{dR}{dP}=Q$.
I also know that the revenue function can be expressed as $R(Q)=P(Q)times Q$, so does this tell me that, because $P$ is a function of $Q$, I cannot treat it as a constant? What is it about this function that requires me to use implicit differentiation? Thanks in advance!
calculus derivatives
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
$endgroup$
add a comment |
$begingroup$
I have a maths related question and I was hoping that you could help me out. I'm trying to understand why, when differentiating the total revenue function, one is required to differentiate implicitly.
The total revenue formula is $R=PQ$, where $R$ is total revenue, $P$ is the unit price and $Q$ is the quantity. I know how to differentiate implicitly... $frac{dR}{dP}=Q+Pfrac{dQ}{dP}$ ... but I don't quite understand why it's needed. That is, why don't we treat $Q$ like a constant, giving $frac{dR}{dP}=Q$.
I also know that the revenue function can be expressed as $R(Q)=P(Q)times Q$, so does this tell me that, because $P$ is a function of $Q$, I cannot treat it as a constant? What is it about this function that requires me to use implicit differentiation? Thanks in advance!
calculus derivatives
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
$endgroup$
3
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
1
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15
add a comment |
$begingroup$
I have a maths related question and I was hoping that you could help me out. I'm trying to understand why, when differentiating the total revenue function, one is required to differentiate implicitly.
The total revenue formula is $R=PQ$, where $R$ is total revenue, $P$ is the unit price and $Q$ is the quantity. I know how to differentiate implicitly... $frac{dR}{dP}=Q+Pfrac{dQ}{dP}$ ... but I don't quite understand why it's needed. That is, why don't we treat $Q$ like a constant, giving $frac{dR}{dP}=Q$.
I also know that the revenue function can be expressed as $R(Q)=P(Q)times Q$, so does this tell me that, because $P$ is a function of $Q$, I cannot treat it as a constant? What is it about this function that requires me to use implicit differentiation? Thanks in advance!
calculus derivatives
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
$endgroup$
I have a maths related question and I was hoping that you could help me out. I'm trying to understand why, when differentiating the total revenue function, one is required to differentiate implicitly.
The total revenue formula is $R=PQ$, where $R$ is total revenue, $P$ is the unit price and $Q$ is the quantity. I know how to differentiate implicitly... $frac{dR}{dP}=Q+Pfrac{dQ}{dP}$ ... but I don't quite understand why it's needed. That is, why don't we treat $Q$ like a constant, giving $frac{dR}{dP}=Q$.
I also know that the revenue function can be expressed as $R(Q)=P(Q)times Q$, so does this tell me that, because $P$ is a function of $Q$, I cannot treat it as a constant? What is it about this function that requires me to use implicit differentiation? Thanks in advance!
calculus derivatives
calculus derivatives
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
asked Nov 18 '15 at 4:40
Rock and a hard placeRock and a hard place
61
61
3
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
1
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15
add a comment |
3
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
1
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15
3
3
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
1
1
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
frac{dg}{dt} = frac{partial g}{partial x} frac{dx}{dt} + frac{partial g}{partial y} frac{dy}{dt} tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $frac{dx}{dt}=1$ and (3) becomes
$$
frac{dg}{dx} = frac{partial g}{partial x} + frac{partial g}{partial y} frac{dy}{dx} tag{34}
$$
it is usually the distinction between $frac{partial g}{partial x}$ and $ frac{dg}{dx}$ which causes difficulty for beginners.
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1534701%2fderivative-of-total-revenue%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
frac{dg}{dt} = frac{partial g}{partial x} frac{dx}{dt} + frac{partial g}{partial y} frac{dy}{dt} tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $frac{dx}{dt}=1$ and (3) becomes
$$
frac{dg}{dx} = frac{partial g}{partial x} + frac{partial g}{partial y} frac{dy}{dx} tag{34}
$$
it is usually the distinction between $frac{partial g}{partial x}$ and $ frac{dg}{dx}$ which causes difficulty for beginners.
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
add a comment |
$begingroup$
really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
frac{dg}{dt} = frac{partial g}{partial x} frac{dx}{dt} + frac{partial g}{partial y} frac{dy}{dt} tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $frac{dx}{dt}=1$ and (3) becomes
$$
frac{dg}{dx} = frac{partial g}{partial x} + frac{partial g}{partial y} frac{dy}{dx} tag{34}
$$
it is usually the distinction between $frac{partial g}{partial x}$ and $ frac{dg}{dx}$ which causes difficulty for beginners.
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
$endgroup$
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
add a comment |
$begingroup$
really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
frac{dg}{dt} = frac{partial g}{partial x} frac{dx}{dt} + frac{partial g}{partial y} frac{dy}{dt} tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $frac{dx}{dt}=1$ and (3) becomes
$$
frac{dg}{dx} = frac{partial g}{partial x} + frac{partial g}{partial y} frac{dy}{dx} tag{34}
$$
it is usually the distinction between $frac{partial g}{partial x}$ and $ frac{dg}{dx}$ which causes difficulty for beginners.
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
$endgroup$
really this is just differentiation, so no need to get too confused here. many relations encountered in science can be expressed in the form:
$$
y = f(x) tag{1}
$$
however a more symmetrical expression is sometimes more appropriate;
$$
g(x,y) = 0 tag{2}
$$
it is trivial (though un-necessary) to rewrite a relation in form 1 as a relation of the second type, viz:
$$
h(x,y) = f(x)-y = 0
$$
however it may not be so simple to translate form 2 into form 1, and often this can only be done under certain restrictions.
using form 2, suppose $x$ and $y$ are both functions (in form 1) of a parameter $t$.
then differentiation by $t$ gives:
$$
frac{dg}{dt} = frac{partial g}{partial x} frac{dx}{dt} + frac{partial g}{partial y} frac{dy}{dt} tag{3}
$$
a special case of this is when the role of the parameter $t$ is taken by the variable $x$. in this case $frac{dx}{dt}=1$ and (3) becomes
$$
frac{dg}{dx} = frac{partial g}{partial x} + frac{partial g}{partial y} frac{dy}{dx} tag{34}
$$
it is usually the distinction between $frac{partial g}{partial x}$ and $ frac{dg}{dx}$ which causes difficulty for beginners.
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
answered Nov 18 '15 at 5:09
David HoldenDavid Holden
14.8k21224
14.8k21224
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
add a comment |
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
$begingroup$
Thanks for that guys!! That helps a lot :)
$endgroup$
– Rock and a hard place
Nov 21 '15 at 3:45
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1534701%2fderivative-of-total-revenue%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
The quantity sold is a function of the unit price! If you raise the price a lot, you will sell very few items. Conversely, if you drop the price significantly, sales will increase.
$endgroup$
– lulu
Nov 18 '15 at 4:42
$begingroup$
Thanks Lulu! I understand that, but I'm not quite getting why implicit differentiation must be used...for example, if $P$ was not a function of $Q$, would $frac{dR}{dP}=Q$?
$endgroup$
– Rock and a hard place
Nov 18 '15 at 4:50
1
$begingroup$
yes. If $Q$ is not a function of $P$ then $frac {dR}{dP}=Q$. The "implicit" part needn't be critical. In some models, $Q$ might be a perfectly explicit function of $P$. In more subtle models, one doesn't know $Q$ as an explicit function, though you can sometimes infer properties of it though formulas of the type you have written down.
$endgroup$
– lulu
Nov 18 '15 at 4:55
$begingroup$
Ok great, thanks for that! So basically, if I have some function with 2 variables, where one variable is a function of the other... like $f(x)=c(x) + x$ or something, then I have to use implicit differentiation? I hope that makes sense haha
$endgroup$
– Rock and a hard place
Nov 18 '15 at 5:03
$begingroup$
I think you are hung up on the word "implicit". Take your case...in the real world $Q$ is a hopelessly complex function of $P$. Sure you can model it, sure you can say some things about it, but there's no notion that you can write it down in any rigid form. Nonetheless, you can talk about its derivative and how small changes in it might effect other variables. Local modeling tends to work a lot better than global modeling. But, really, this says more about the nature of $Q(P)$ than it does about the process of differentiation.
$endgroup$
– lulu
Nov 18 '15 at 12:15