# Derivative of a function of another function

• I
• kent davidge

#### 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.

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

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.

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

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 $$\frac{\partial f}{\partial \varphi} = \frac{\partial f}{\partial g}\frac{\partial g}{\partial \varphi}$$   ##\dfrac{df}{d \varphi}=\dfrac{df(g)}{dg}\cdot\dfrac{dg(\varphi)}{d\varphi}##.

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)##.

• kent davidge
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.