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I am not too grounded in differentiation but today, I was posed with a supposedly 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 $\delta$? – Zhen Lin Jul 23 '12 at 15:11
Do you mean $x = r \cos \theta$ and $y = r\sin \theta$? – Jesse Madnick Jul 23 '12 at 15:59
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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 $\partial w / \partial r$. – wildildildlife Jul 23 '12 at 16:50

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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However, if $x=x(r,\theta)$ and $y=y(r,\theta)$, then the assumption that $y$ stays constant when $x$ changes is no longer valid.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...' – Chibueze Opata Jul 23 '12 at 15:41
@ChibuezeOpata Since $x$ and $y$ are now functions of $r$ and $\theta$, we cannot assume that $x$ and $y$ are independent. – M Turgeon Jul 23 '12 at 15:42
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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"? – Jesse Madnick Jul 23 '12 at 15:56
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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: en.wikipedia.org/wiki/Total_derivative – Arkamis Jul 23 '12 at 16:04
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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. – Arkamis Jul 23 '12 at 18:59
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