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I am not too grounded in differentiation but today, I was posed with a supposedly easy question $w = f(x,y) = x^2 + y^2$ where $x = r\sin\theta $ and $y = r\cos\theta$ requiring the solution to $\partial w / \partial r$ and $\partial w / \partial \theta $. I simply solved the former using the trig identity $\sin^2 \theta + \cos^2 \theta = 1$, resulting to $\partial w / \partial r = 2r$.

However I was told that this solution could not be applied to this question because I should be solving for the total derivative. I could not find any good resource online to explain clearly to me the difference between a normal derivative and a total derivative and why my solution here was wrong. Is there anyone who could explain the difference to me using a practical example? Thanks!

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    What is $w$? Did you mean $\partial$ instead of $\delta$?2012-07-23
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    Do you mean $x = r \cos \theta$ and $y = r\sin \theta$?2012-07-23
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    I agree: since $w(r,\theta)=r^2$, I would say $\partial w / \partial r=2r$. If I were you, I would ask that person for a precise definition of 'total derivative' and of $\partial w / \partial r$.2012-07-23
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    OP: Did they say that your answer was wrong, or only that you solved it using the wrong method?2013-12-02
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    @NeilTraft If I remember correctly, -- that the answer was wrong because the question was asking for the *total derivative*.2015-02-24
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    Do you know what the answer was? I keep getting $2r$ even when using Arkamis' formula.2015-02-25
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    Given that $r$ and $\theta$ are independent , I think your solution is totally right.2015-05-23
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    I'm really confused - when I write a total derivative, it looks like $d/dr$ and when I write a partial, it looks like $\partial / \partial r$. If I was asked what $\partial w/ \partial r$ and I didn't see an $r$, I'd say 0 and laugh. Clearly, $dw/dr = 2 r$. Are either of these wrong? If not, why is the OP asking about the partial, answering with the total and then being told to find the total instead?2015-05-28

6 Answers 6

50

The key difference is that when you take a partial derivative, you operate under a sort of assumption that you hold one variable fixed while the other changes. When computing a total derivative, you allow changes in one variable to affect the other.

So, for instance, if you have $f(x,y) = 2x+3y$, then when you compute the partial derivative $\frac{\partial f}{\partial x}$, you temporarily assume $y$ constant and treat it as such, yielding $\frac{\partial f}{\partial x} = 2 + \frac{\partial (3y)}{\partial x} = 2 + 0 = 2$.

However, if $x=x(r,\theta)$ and $y=y(r,\theta)$, then the assumption that $y$ stays constant when $x$ changes is no longer valid. Since $x = x(r,\theta)$, then if $x$ changes, this implies that at least one of $r$ or $\theta$ change. And if $r$ or $\theta$ change, then $y$ changes. And if $y$ changes, then obviously it has some sort of effect on the derivative and we can no longer assume it to be equal to zero.

In your example, you are given $f(x,y) = x^2+y^2$, but what you really have is the following:

$f(x,y) = f(x(r,\theta),y(r,\theta))$.

So if you compute $\frac{\partial f}{\partial x}$, you cannot assume that the change in $x$ computed in this derivative has no effect on a change in $y$.

What you need to compute instead is $\frac{\rm{d} f}{\rm{d}\theta}$ and $\frac{\rm{d} f}{\rm{d} r}$, the first of which can be computed as:

$\frac{\rm{d} f}{\rm{d}\theta} = \frac{\partial f}{\partial \theta} + \frac{\partial f}{\partial x}\frac{\rm{d} x}{\rm{d} \theta} + \frac{\partial f}{\partial y}\frac{\rm{d} y}{\rm{d} \theta}$

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    I am confused at: 'so if you compute $\partial f / \partial x$, you cannot assume that the change in x computed has no effect on a change in y...'2012-07-23
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    @ChibuezeOpata Since $x$ and $y$ are now functions of $r$ and $\theta$, we cannot assume that $x$ and $y$ are independent.2012-07-23
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    Right. So, $x = x(r,\theta)$ and $y = y(r,\theta)$. This means that when we compute the partial derivative $\partial f/\partial x$, we cannot assume that $y$ is being held constant, because if $x$ changes, it is because either $r$ or $\theta$ changes (or both). And if $r$ or $\theta$ change, then $y$ changes. So $\partial (y^2)/\partial x$ is not zero. Remember that differentiation is an operation taken in the limit of small changes to an independent variable, so $\partial f/\partial x$ implies that $x$ is undergoing some small change.2012-07-23
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    I am also confused by this post. Where does the formula $\frac{df}{d\theta} = \frac{\partial f}{\partial \theta} + \frac{\partial f}{\partial x}\frac{dx}{d\theta} + \frac{\partial f}{\partial y}\frac{dy}{d\theta}$ come from? (In particular, why are there three terms on the right-hand side?) And what is the exact definition of "total derivative"?2012-07-23
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    It comes from the chain rule and letting $x$ and $y$ be functions of $r$ and $\theta$. Technically, I should have probably used partial symbols in $\rm{d} x/\rm{d}\theta$. A "total derivative" can be thought of as the computation of the derivative of a parametric function with respect to the parameter(s). See more here: http://en.wikipedia.org/wiki/Total_derivative2012-07-23
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    I see. So you're regarding $f$ as a function $f(\theta, x, y)$, then writing $df = \frac{\partial f}{\partial \theta}d\theta + \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$, and then dividing by $d\theta$.2012-07-23
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    Yes, in a sense. And in the OP's question, you could add an $r$: $f(r,\theta,x,y)$, but this is not how I prefer to write the function definition, as it hides the fact that the total derivative of $f(x=r\cos\theta,y=r\sin\theta)$ is simply the chain rule applied to find $\partial f/\partial \theta$.2012-07-23
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    @Ed: But what does $\frac{\partial f}{\partial \theta}$ even mean if $f$ is not a function of $\theta$ directly? I would expect that any effect of a change in $\theta$ (for fixed $r$) was taken care of by the chain rule's $\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}$. Can you provide a worked example where you show what $\frac{\partial f}{\partial \theta}$ is, concretely?2012-07-23
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    In this particular case, $f$ is not a function of $\theta$ directly, and consequently $\partial f/\partial \theta = 0$. I will try to post a worked example in my answer when I have some more time later.2012-07-23
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    In that case, I think it is misleading to suggest to the OP that he needs something more than just the chain rule in his situation.2012-07-23
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    The OP asked what the "total derivative" means. In one regard you could say that the "total derivative" is nothing more than applying the chain rule in such a way that you "end up" with derivatives with respect to only the parameter. However, contextually, computing the "total derivative" means something different than just applying the chain rule. For instance, $f(x,y) = xy$ can be computed using the chain rule, but it may not be the total derivative if $x=x(t)$ and $y=y(t)$. So, in other words, the total derivative applies the chain rule, but it means something slightly stronger.2012-07-23
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    How would you use the chain rule to compute $f(x,y)=xy$? Evaluating that function is just a matter of multiplying two numbers. And there is still no explanation of what "total derivative" actually _means_ here, except the sum of the ordinary chain rule result plus a mysterious $\partial f/\partial \theta$ term that has not been defined except saying that it is $0$ in the example at hand. The Wikipedia article describes a situation where we have not only $f$ but _also_ a particular subset of $\mathbb R^2$ that $(x,y)$ is constrained to, but there's no such constraint in the question here.2012-07-23
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    I don't understand how the term $\partial d/\partial \theta$ is mysterious at all. The total derivative represents the derivative of a function with respect to parametric variables. If you have $f(x,y) = x^2+y^2+\theta$, then the $\partial f/\partial \theta$ is just 1. The total derivative has numerous analogs, such as the material derivative in fluid mechanics.2012-07-24
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    The total derivative represents the net effect of the change in the function $f$ due to a small change in some underlying variable $\theta$. When you write $f$ in terms of other variables $x$ and $y$, that depend on $\theta$, then it is *meaningless* to compute $\partial f/\partial x$ unless you *also* consider the fact that $x$ only changes due to an underlying change in $\theta$. There is, however, nothing that says that $f$ has to depend explicitly on $\theta$.2012-07-24
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    To summarize quite succinctly from Wikipedia: "Calculation of the total derivative of $f$ with respect to $t$ does not assume that the other arguments are constant while $t$ varies; instead, it allows the other arguments to depend on $t$. The total derivative adds in these indirect dependencies to find the overall dependency of $f$ on $t$.2012-07-24
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    @Emily An issue I see with this approach is that the given function $w=f(x(r, \theta) , y(r, \theta))$ does not depend on $\theta$ explicitly. Forcing $f$ to depend on $\theta$ explicitly with $f(\theta, x(r,\theta), y(r,\theta) \implies \frac{\rm{d} f}{\rm{d}\theta} = \frac{\partial f}{\partial \theta} + \frac{\partial f}{\partial x}\frac{\rm{d} x}{\rm{d} \theta} + \frac{\partial f}{\partial y}\frac{\rm{d} y}{\rm{d} \theta}$ , but $ \frac{\partial f}{\partial \theta}=0 $ This is discussed here https://en.wikipedia.org/wiki/Total_derivative#Differentiation_with_indirect_dependencies2017-11-05
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    Sorry for the necropost, but, how does one take the total derivative of a total derivative?2018-09-04
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    $f$ is not a direct function of $\theta$, rather $\theta$ works as a parameter. So in the final eqn where you say "$\frac{\partial f}{\partial \theta}+...$", (chain rule), that particular term should be removed. You might say, that the partial derivative reduces to zero, but this form might look a bit odd. The rest of the answer is clear.2018-10-08
23

I know this answer is incredibly delayed; but just to summarise the last post:

If I gave you the function

$$ f(x,y) = \sin(x)+3y^2$$

and asked you for the partial derivative with respect to $x$, you should write:

$$ \frac{\partial f(x,y)}{\partial x} = \cos(x)+0$$

since $y$ is effectively a constant with respect to $x$. In other words, substituting a value for $y$ has no effect on $x$. However, if I asked you for the total derivative with respect to $x$, you should write:

$$\frac{df(x,y)}{dx}=\cos(x)\cdot {dx\over dx} + 6y\cdot {dy\over dx}$$

Of course I've utilized the chain rule in the bottom case. You wouldn't write $dx\over dx$ in practice since it's just $1$, but you need to realise that it is there :)

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    I hope you did notice that my answer was for $\partial w / \partial r$2013-12-13
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    Helpful approach. Can you please compare/contrast the total derivative w.r.t. x (which you show) with the total derivative w.r.t. y (which you did not show)?2017-05-30
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    $\frac{df(x,y)}{dy}=\frac{\partial f} { \partial x } \frac{dy}{dx} + \frac{\partial f }{\partial y } \frac{dy}{dy}= \cos x \frac{dy}{dx} + 6y (1) $2017-11-05
16

Does everyone agree that the poster arrived at the correct answer?

People write $$\frac{\partial}{\partial t}g(x(t),t)$$ or $$\frac{\text{d}}{\text{d} t}g(x(t),t)$$

The first is typically used to mean "the derivative of function $g$ with respect to the second argument". The second usually means the "total derivative". There are variations on this. Some people omit the arguments and just write, for example, $\frac{\partial}{\partial t}g$

So for example: if $x$ is secretly a function of $t$, then the notation $\frac{d}{dt}f(x,t)$ is called the total derivative and is an abbreviation for the (single-variable derivative) $g′(t)$ where $g(t)=f(x(t),t)$. In applying the chain rule to the last expression, you would need some way to denote "the derivative of f with respect to its first argument" many people would write $\frac{\partial}{\partial x}f$ for this, but in many cases this is confusing as I explain in the example below.

The wide-spread math notation here confuses many people and I think it is pretty much unnecessary to use it. If you want to take a total derivative, construct explicitly the function (like $g$ above) and take a single-variable derivative. Otherwise, the explanations for the difference between total and partial derivatives needs you to make appeals like temporarily fixing variables or saying that a variable is effectively constant or switching between thinking of $x$ as a function and as an expression. These are all fuzzy things you can do successfully once you already feel comfortable with what's going on. But otherwise, it pays to think carefully about what's really happening.

Your example

The problem stems from the conflation of an expression and a function. You did this when you wrote $w = f(x,y) = x^2 + y^2$. In that case, many will write

$\frac{\partial}{\partial x}w$ and

$\frac{\partial}{\partial x} f(x,y)$

(which are equivalent). This sort of makes sense. In both cases, the thing to the right of the differential operator is an expression which contains $x$ and $y$. The thing that is produced by applying that operator is also an expression in the same variables. This is also true of what $\frac{d}{dx}$ means. For the particular expressions above, I would just use that.

The actual purpose of the partial derivative is to take derivatives of functions with respect to one of its arguments, not expressions. That's not what's happening above. That is what's happening when people write:

$\frac{\partial}{\partial x} f$.

$f$ is not an expression. It is a function. I personally do not like this notation. You could have defined an identical $f$ by writing $f(a,b) = a^2 + b^2$. The variables that appear in the definition of a function are, in the strictest sense, invisible to the rest of the world. It's just a convenient way of stating "$f$ is a function that takes two arguments. It squares the first, squares the second, and returns the sum of the squares". Instead of having to write that sentence out (which people had to do before inventing better notation), you can instead give names to the arguments of $f$ so that you can easily refer to them when defining $f$.

But when you write $\frac{\partial}{\partial x} f$, then you are using some knowledge of how you defined $f$---that you chose the name $x$ for the first argument. It can be useful to have names for function arguments instead of just referring to their position (first, second, etc. argument), and so that's why the partial notation survives, but I think the notation needs to improve for this.

What someone typically means when they write $\frac{\partial}{\partial x} f$ is roughly "the function that takes two arguments and returns the sensitivity of $f$ with respect to its first argument". So if you're at some point $(a,b)$ or $(x,y)$ or whatever, and you wiggle the first argument $a$ or $x$, how much does the output of $f$ wiggle? That is the question that the gradient of a function is supposed to answer. This is probably what someone means if they say "normal derivative" They are thinking about only a single function, with possibly multiple arguments. And they are trying to make an object that tells you how sensitive the output of the function is to a change in each of the inputs.

The total derivative usually means that somewhere you've implicitly defined some new functions. In this case, you have made functions $x(r,\theta) = r \sin(\theta)$ and $y(r,\theta) = r \cos(\theta)$, and you can compose these functions, making a new function: $$g(r,\theta) = f(x(r,\theta),y(r,\theta))$$

Notice again, that $r$ and $\theta$ are chosen only to give a human information about connotation of this function. If we processed things purely symbolically, then the definition of $g$ could as well have been

$$g(input_1,input_2) = f(x(input_1,input_2),y(input_1,input_2))$$

And so when the problem asked you to find $\frac{\partial}{\partial r} w$, there are two, in the end identical, interpretation of what that means. Either construct the function $g$ as I did above, and report its sensitivity with respect to the first argument. OR substitute the expressions for $x$ and $y$ into the expression for $w$. Now you have an expression for $w$ in terms of $r$ and $\theta$. I prefer the approach that thinks about functions. This is how we organize code and I think this is how we should organize math. When you deal with expressions, you effectively have a ton of global variables.

So how do we compute $\partial_1 g$, which is just the notation for "make a function with the same arity (number of inputs) as $g$, such that it evaluates the the derivative of the function $g$ with respect to its first argument"? It's just the chain rule.

$$[\partial_1 g](r,\theta) = [\partial_1 f](x(r,\theta), y(r,\theta)) \cdot [\partial_1 x](r,\theta) + [\partial_2 f](x(r,\theta), y(r,\theta)) \cdot [\partial_1 y](r,\theta)$$

We can see why thinking about things in this way is not popular! But this is the clearest, most mechanical, way to think about it. Otherwise you are relying on implicit punning of $x$ as a function and as an expression. Choose one and stick with it!

Anyway, to simplify the above definition, which didn't care about the definitions of $f$, $x$, or $y$, we need to use the definitions.

$f(x,y) = x^2 + y^2$ and therefore

  • $[\partial_1 f](x,y) = 2x$
  • $[\partial_2 f](x,y) = 2y$

$x(r,\theta) = r\sin(\theta)$ and therefore

  • $[\partial_1 x](r,\theta) = \sin(\theta)$

likewise

  • $[\partial_1 y](r,\theta) = \cos(\theta)$

FURTHERMORE, though we don't need it at the moment

  • $[\partial_2 x](r,\theta) = r\cdot \cos(\theta)$
  • $[\partial_2 y](r,\theta) = -r\cdot \sin(\theta)$

So again, the function is

$$[\partial_1 g](r,\theta) = [\partial_1 f](x(r,\theta), y(r,\theta)) \cdot [\partial_1 x](r,\theta) + [\partial_2 f](x(r,\theta), y(r,\theta)) \cdot [\partial_1 y](r,\theta)$$

substituting the functions we just computed:

$$[\partial_1 g](r,\theta) = 2x(r,\theta) \cdot \sin(\theta) + 2y(r,\theta) \cdot \cos(\theta)$$

and substituting $x$ and $y$

$$[\partial_1 g](r,\theta) = 2r\sin(\theta) \cdot \sin(\theta) + 2r\cos(\theta) \cdot \cos(\theta)$$

which, after using the very trig identity you used, is

$$[\partial_1 g](r,\theta) = 2r$$

Yet another way to make the same point:

When you see the notation $g'(x)$, you can group that as $[g'](x)$. You've made a new function, called "g prime", which is the derivative of $g$, and you're evaluating it at point $x$. $g'(y)$ means the same thing, except you're evaluating at the point $y$. The multidimensional analogue of this is $\nabla g(\mathbf{x})$. You should parse that as $[\nabla g](\mathbf{x})$.

This is not the case with the notation $\frac{d}{dx} g(x)$. If you parse that as $[\frac{d}{dx} g](x)$, you get confused because what does $x$ mean in the scope of the brackets? You don't have to give meaning to it because it should be meaningless. The operator $\frac{d}{dx}$ applies to an expression, not a function.

But, what people will routinely do is define

$g(x)= x^2+sin(x)+\text{whatever expression in }x$

and then write $\frac{d}{dx} g(y)$ when they really should have written $g'(y)$. They don't do this very often in the single-variable case, but they do it in the multi-variable case. I just showed the single-variable case because it's clearer to see the problem with it.

My inspiration for this answer comes from http://groups.csail.mit.edu/mac/users/gjs/6946/sicm-html/book-Z-H-78.html#%_sec_Temp_453)

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    I'm having trouble finding a clear definition of the term "total derivative" in all that text. Could you perhaps make it stand out more?2015-05-22
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    I think one conclusion from the wall-o-text is that the concept of "total derivative" is due to sloppiness in math notation, and that we are better off without it. As such, I don't think there IS a clear definition. Do you see what I mean?2015-05-22
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    x @Gus: No, I don't see what you mean. If you want to abolish the concept that's fine, but if you want me to _agree_ that the concept should be abolished, I'll have to know what the concept we're going to abolish IS before I can agree to it.2015-05-22
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    In my original answer, I have "the total derivative usually means..." Do you think I need to make this stand out typographically?2015-05-22
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    @HenningMakholm, I clarified and reorganized. I hope the point is clearer now! Since you are CS/PL person, I think you can help me make this even clearer. What I'm really getting at is some distinction between a function and an expression and named arguments versus positional arguments. People get confused because of a scoping issue. in the expression ∂f(x,y)/∂x, the x in the denominator is named argument of f. The x in the numerator is something else: a variable defined in the current scope.2015-05-23
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    x @Gus: What I think there is still missing in the answer is a _definition of the term "total deriviative"_. The closest it gets is a sentence that says "the total derivative usually ***means*** that (someone did such and such)". There should be a simple declarative sentence somewhere saying "The total derivative ***is*** (a clear description of the thing denoted by the term)". You're still assuming that the reader already knows the meaning of the term and jumping directly to critiquing the reasons one might have for using it.2015-05-23
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    What I'm suspecting from your text is that this definition would be something like: "If $x$ and $y$ are secretly functions of $z$, then the notation $\frac{df(x,y)}{dz}$ is called a **total derivative** and is an abbreviation for $\frac{\partial}{\partial t} f(x(z),y(z))$. The reason why ordinary derivatives use a $\partial$ sign instead of a $d$ is to avoid confusion with the total-derivative notation".2015-05-23
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    If $x$ is secretly a function of $t$, then the notation $\frac{\text{d}}{\text{d}t} f(x,t)$ is called the *total derivative* and is an abbreviation for the (single-variable derivative) $g'(t)$ where $g(t)=f(x(t),t)$. In applying the chain rule to the last expression, you would need some way to denote "the derivative of $f$ with respect to its first argument" many people would write $\frac{\partial}{\partial x} f$ for this, but in many cases this is confusing.2015-05-23
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    This is the best response but not a very good answer and I wish there was a way to mark it as the best detailed response to the issue. I could not completely comprehend it at the time you posted it but thanks.2017-03-30
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    @ChibuezeOpata don't worry about marking it as the best detailed response. I'm glad it was helpful, if only years later. If you can think of how to make it clearer, please edit the answer or let me know.2017-03-31
1

Partial derivative is the derivative of a function with several independent variables with respect to any one of them, keeping the others constant.

The symbols $ \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y} $ are used to denote such differentiations.

And the expressions $ \dfrac{\partial u}{\partial x}, \dfrac{\partial u}{\partial y} $ are called partial differential coefficients of $u$ with respect to $x$ and $y$.

So, if $u=f(x,y)$, then $ \dfrac{\partial u}{\partial x} $ can be calculated by differentiating $u$ with respect to $x,$ keeping $y$ constant.

While in case of Total derivative, we don't assume the other variables to be constant, their change with respect to the change in that variable is also taken into consideration.

So, if $u=f(x,y)$, where $x=\phi_1(t)$, and $y=\phi_2 (t)$, it can be calculated as:

$ \dfrac{du}{dt} =$ $ \dfrac{\partial u}{\partial x}. $ $ \dfrac{dx}{dt} $ + $ \dfrac{\partial u}{\partial y}. $ $ \dfrac{dy}{dt} $

So, if $x$ and $y$ both depend on $t$, then change in $x$ will lead to change in $t$ which in turn will lead to change in $x$ (since $y$ is also a function of $t$), hence the formula mentioned above.

In your question, Since $x$ and $y$ both depend on $\theta$ (Assuming $r$ is a constant), so in order to find total derivative of $f(x,y)$ with respect to $\theta$ , apart from the partial derivatives of $u$ with respect to $x$ and $y$ you will need to consider change in $x$ as well as change in $y$ with respect to $\theta$.

Summarizing the whole thing:

Total derivative is a measure of the change of all variable, while Partial derivative is a measure of the change of a particular variable having others kept constant.

Hope this helps!

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    This doesn't seem to answer the question. The question was not what is a partial derivative (everyone who's ever taken a basic calculus course knows that), but that is a "total derivative", a mysterious concept which nobody seems to be able to give a straight definition for.2015-05-22
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    The expression in the end is for total derivative only. @HenningMakholm2015-05-22
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    x @Nivedita: But the OP asked to understand what "total derivative" _means_, not just get an expression for his particular situation without any explanation of its significance, or even _what makes that expression right_.2015-05-22
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    @HenningMakholm I have edited the answer. Let me know if more explanation is needed.2015-05-22
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It may be easier to imagine a figure with orthogonal x and y co-ordinates for the base and a functional result (i.e. some function, w, of two variables x and y) plotted as a surface on the vertical z axis. If we look at the result for a change in the function that we obtain when we keep y constant and let x vary, it is a tangent to the surface of slice taken through that surface parallel to the x axis. Of course you get an equivalent picture for letting y vary but keeping x constant. Now imagine a change in the function but we are letting both x and y vary simultaneously, delta w is the change in w and basically we sum the changes in the function to get delta w, del w =f(x + delx, y + del y) - f(x,y) if we expand for del w and go to the limit we get for dw = the partial derivative wrt x times dx plus the partial derivative wrt y times dy. If x and y are both functions of a single variable t, then so is w, and we can divide each term by dt which is the total derivative wrt to t of the function.

This is the classic example of the basic concepts and you can find a version of it here:-

https://www.math.uwaterloo.ca/~ahamadeh/math217_p2.pdf

For a good illustrative example I like the rate of change of an expanding cylinder volume, V is given by V= (PI)(r^2)(h), r = radius h = height, now use the previous idea expression for delta w, but here its delta V, divide both sides by delta t and then let delta t go to zero.

It only gets slightly more complicated when we use the methodology to find differential coefficients of implicit functions, but we use similar methods. Textbook examples often let z stand for the function of x and y, then you just form delta z (as 'normal') divide both sides by delta x and let delta x go to zero giving an expression for dz/dx. Often you are given info about z, may be z = 0 (constant) so dz/dx = 0.

If you tack on to these ideas the change of variable idea, which is a bit more of the same really, e.g. say z is a function of x and y z = f(x,y) and x and y are in turn functions of two other variables u and v , then z is a function of u and v, so you form delta z as 'normal' in terms the partial diff of z wrt x times delta x plus the partial diff of y times delta y, divide both sides by delta u and let delta u got to zero, v is being kept constant for the time being. That gives you the partial differential of z wrt u, and you follow the same procedure to get an expression for the partial differential of z wrt v .

With these you have most of the basic tools to handle such problems, and I suppose the basic answer to your question is that you have an implicit function, so when you want the change in the function when x is varying you have to add on the 'extra' bit. The concept pops up in a number of places, may be your function describes the temperature of a volume element, its cooling with time, but its also moving and the spatial co-ordinates bring it near a heat source, so you have to add the two effects. So if you are not careful you miss that 'extra' spatial term and just have the 'pure' time term.

Apologies for the wordiness of this reply, but just maybe a half reasonable idea of what is going on has been conveyed (for time I'd imagine the whole function surface altering shape with time in 3 D perhaps we can only have snapshots in time.).

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I find some of the answers (and comments) above to be a bit confusing. I want to address some of the issues brought up. The original question of the OP was to find the total derivative $ \frac{dw}{dr} $ for the function: $$w=f(x,y)= x^2 + y^2,~~ x=r \sin \theta ,~~y = r \cos \theta $$ assuming that $r, \theta $ are independent variables.

On the face of it finding $ \frac{dw}{dr} $ is not possible if $r, \theta $ are independent of each other.

It is true that $$ \frac{\partial w}{\partial r }= 2r$$ Proof: $$ \frac{\partial w}{\partial r } = \frac{\partial w}{\partial x } \frac{\partial x }{\partial r} + \frac{\partial w}{\partial y } \frac{\partial y }{\partial r} $$ Plugging in $$ \frac{\partial w}{\partial r } = 2x~ ( \sin \theta ) + 2y ( \cos \theta ) $$ Substituting using the given $x $ and $y $ equations $$\frac{\partial w}{\partial r } = 2( r \sin \theta ) ~ ( \sin \theta ) + 2( r \cos \theta ) ( \cos \theta ) = 2r ( \sin^2 \theta + \cos^2 \theta ) = 2r $$

We can relax the assumption that $r$ and $\theta $ are independent of each other to find $ \frac{dw}{dr} $. The computation is quite a bit more involved. We would have to temporarily assume that $\theta$ is a function of $r$. $$ \frac{dw}{dr} =\frac{\partial w}{\partial x } \frac{\partial x}{\partial r } \frac{dr}{dr }+ \frac{\partial w}{\partial x } \frac{\partial x}{\partial \theta }\frac{d\theta }{ dr }+\frac{\partial w}{\partial y } \frac{\partial y}{\partial r } \frac{dr}{dr }+ \frac{\partial w}{\partial y } \frac{\partial y}{\partial \theta }\frac{d\theta }{ dr } $$ It is true that an early substitution gives us $$ w= ( r \sin \theta) ^2 + ( r \cos \theta)^2 = r^2$$ But it would be misleading to state that $$ \frac{dw}{dr}= 2r $$ since $$ w = f( r, \theta) $$ An analogous scenario is found when we want the slope at a point on the surface $$ z=f(x,y) = x^2$$ We would still use partial derivative even though we have $$ \frac{\partial z}{\partial x } = \frac{dz}{dx} = 2x$$ We can be more explicit and define $$z=f(x,y) = x^2 + 0y$$