Derivative of a function of another function

[kent davidge]

This is really a simple question, but I'm stuck.

Suppose we have a function $\vartheta'(\vartheta) = \vartheta$ and that $\vartheta = \vartheta(\varphi)$ and we know what $\vartheta(\varphi)$ is. How should I view $\frac{\partial \vartheta'}{\partial \varphi}$? Should I set it equal to zero because $\varphi$ does not appear explicitely? But as I said, I know what $\vartheta(\varphi)$ is in this particular case, and it is not even linear in $\varphi$, so it doesn't make sense to say that its derivatives vanish.

Edit: I added a crucial correction to my post.

 

[fresh_42]

What does $\vartheta '( \vartheta )$ mean?

 

[kent davidge]

What does $\vartheta '( \vartheta )$ mean?

I'm sorry, that was a bad choice of notation. $\vartheta'$ is simply a function, the prime is there to differentiate it from $\vartheta$. It does not mean derivative. If you prefer, let's call it $\Theta$ from now on to cause no confusion.

 

[zinq]

Still truly bad notation, especially because a prime superscript placed on the symbol for a function typically means its derivative. And this question is about derivatives.

 

[kent davidge]

Still truly bad notation, especially because a prime superscript placed on the symbol for a function typically means its derivative. And this question is about derivatives.

 

[fresh_42]

So we are talking about the chain rule? $g=g(\varphi)$ and $f=f(g)$, so $f=f(g(\varphi))$?

 

[kent davidge]

So we are talking about the chain rule? $g=g(\varphi)$ and $f=f(g)$, so $f=f(g(\varphi))$?

oh yes. And I would like to know $$\frac{\partial f}{\partial \varphi}$$ If $\varphi$ were the argument of $f$ directly, then it would be the same as the total derivative of $f$ wrt it. But since $\varphi$ is the argument of $g$, I'm not sure how to proceed.

chain rule?

oops, this just reminded me that we can use the chain rule also with partial derivatives. so $$\frac{\partial f}{\partial \varphi} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial \varphi}$$

 

[fresh_42]

$\dfrac{df}{d \varphi}=\dfrac{df(g)}{dg}\cdot\dfrac{dg(\varphi)}{d\varphi}$.

 

[Stephen Tashi]

Let's try different notation for the OP and use an example.

Suppose we are given $f(y) = 2y$ and $y = x^2$. Then what is $\frac {df}{dx}$?

Is it zero because $x$ does not appear in the formula $f(y) = 2y$ ? Or is it $4x$ because $f(x) = 2(x^2)$ ?

This involves ambiguous and bad notation - and notation that is customary, especially in writing about physics! Beginning students in mathematics are taught that a function has a definite domain and co-domain. So if two functions have different co-domains, they must be different functions. The function $f(y)$ can take on negative values. The function $f(x)$ cannot. So we shouldn't use the name $f$ for both functions. However, it is common practice to do so.

When we are "given" that $y = x^2$, it isn't clear how the name $f$ should be used in representing this fact. To be unambiguous, we could say there is function $f(y) = 2y$ and another function $B(x) = f(x^2) = 2x^2$.

Technically speaking, if $x$ denotes a variable not in the domain of $f$ then $\frac{df}{dx}$ is undefined. However, convention says that we interpret $\frac{df}{dx}$ by pretending $f(y)$ is actually a function of two variables $(x,y)$ that is constant with respect to $x$. With that convention, $\frac{df}{dx} = \frac{\partial f}{\partial_x} = 0$.

I'd say the most common interpretation of the above example in physics is that $\frac{df}{dx} = 0$.

In a math text, the author might want to give students an exercise in the chain rule. He would expect them to interpret $\frac{df}{dx}$ as the derivative of the function $B(x)$.

 

[Mark44]

Suppose we have a function $\vartheta'(\vartheta) = \vartheta$ and that $\vartheta = \vartheta(\varphi)$ and we know what $\vartheta(\varphi)$ is.

Why are you using Greek letters here? The examples shown in the posts by @fresh_42 and @Stephen Tashi are much clearer, not to mention easier to type.

 

[kent davidge]

Why are you using Greek letters here? The examples shown in the posts by @fresh_42 and @Stephen Tashi are much clearer, not to mention easier to type.

Indeed... As if it wasn't enough not understanding the topic, I come up with this notation to make things more difficult.