# I Partial derivatives method

1. Aug 1, 2017

### dyn

Hi.
If I have a function f ( x , t ) = x - 6t with x ( t ) = t2 and I take the partial derivative of f with respect to x I get the answer 1 as t acts as a constant so its derivative is zero. But if I substitute t with x1/2 I get the answer 1 - 3x-1/2 which is obviously different and wrong , I think ! So , why can't I do this method ?
Thanks

2. Aug 1, 2017

### FactChecker

If t is constant, then so is x=t2. Better is t = x0.5 and df/dx = 1 - 6*0.5*x-0.5

3. Aug 1, 2017

### dyn

with f(x,t) = x-6t and x(t ) = t2 then ∂f/∂x = 1 but if I substitute t for x1/2 I get a different answer. I think my first answer is correct but I don't understand why it doesn't work if I make the substitution ?

4. Aug 1, 2017

### FactChecker

No, your first answer is wrong. You can't say that x = t2, that t is constant, and that x changes (since dx ≠ 0 ). Since x and t are related, you have to say that df/dx = dx/dx -6 dt/dx and figure it out from there. t is not constant.

5. Aug 2, 2017

### Dragon27

Well, when you take a partial derivative of a function of several variables, what you get will depend, of course, on the dependencies you decide to impose on these variables. For example, you can just use a partial derivative chain rule:
$$\frac{\partial f(u,v)}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$$
Now you'll have to decide how these variables (abstract first and second variables of the function of two variables $f$; I even made a different notation for these variables - $u$ and $v$, to underscore that) depend on your $x$. Since $u$ is equal to $x$, then $\frac{\partial u}{\partial x} = 1$, and you only have to decide whether you take $v=t$ independent of $x$ (then you'll basically get a partial derivative of $f$ with respect to the first variable, since $\frac{\partial u}{\partial x} = \frac{\partial t}{\partial x} = 0$) or not:
$$\frac{\partial f}{\partial u} + \frac{\partial f}{\partial t}\frac{\partial t}{\partial x}$$

6. Aug 2, 2017

### dyn

I will quote the full question and answer
Calculate ∂f/∂x for f(x,y,t) = 3x2 +2xy + y1/2t - 5xt . With x(t) = t3 and y(t) = 2t5
The answer is given as ∂f/∂x = 6x + 2y -5t.
This looks to me like differentiating with respect to x while holding y and t constant even though x depends on t

7. Aug 2, 2017

### Dragon27

Probably because by $\partial f / \partial x$ they meant a partial derivative of the function $f$ with respect to the first argument.
Now that I think of it, it would make more sense to call the other case (differentiate $f$ over all of its variables with the use of the chain rule) taking a full derivative with respect to $x$ (when we look at $f$ as a function over $x$ only).

8. Aug 2, 2017

### FactChecker

Obviously this is a different example, but it certainly looks like you are correct about what they are doing. I don't think that I agree, but maybe I am missing a trick. Is there any more explanation or context given for this example? What subject is this supposed to illustrate? There are techniques for handling constraints where I could understand their partial derivatives.

9. Aug 2, 2017

### dyn

It was just a question I found on partial derivatives. I understand that there is a difference between a partial derivative and a full/total derivative. My understanding is that a partial derivative means you treat all the other variables as constants but now i'm confused if one of the other variables is related to t ie. x(t) = t3

10. Aug 2, 2017

### Dragon27

If we took all dependencies into consideration $f$ would be a function of just one variable, kind of like $f(x,y(x),t(x))$, where each of $y$ and $t$ is itself a function of $x$, and we would have to find their derivatives with respect to $x$. That wouldn't be exactly a partial derivative.

Last edited: Aug 3, 2017
11. Aug 3, 2017

### dyn

I'm getting a bit confused now. Is the stated answer given above correct ? And if it is , why is it correct ?

12. Aug 21, 2017

### Stephen Tashi

Functions have a domain and co-doman. The function $f = f(x,y,t)$ has a domain consisting of triples of real numbers. The function $f$ is not the same function as the function $g = g(t) = 3(t^3)^2 + 2(t^3)(2t^5) + (2t^5)^{1/2}t + 5(t^3)t$, which has a domain consisting of singletons of real numbers.
The question in your example defines the function $f$ and then attempts to confuse us by saying "With $x(t) = t^3$ and $y(t) = 2t^5$". We are tempted to think of "$f$" as denoting two different functions, one whose domain is triples of real numbers and another whose domain is singletons of real numbers. It is not correct mathematical notation to use "$f$" to denote both $f(x,y,t)$ and $f(t) = g(t)$ within the same problem.
Correct mathematical notation does not use the same letter to denote two different functions. However, you will find that articles on physics and other applications of mathematics do use the same letter to denote two (or more!) different functions. In reading such articles, you must determine from the context of the problem being solved which function "$f$" is intended to represent. This is discussed in the Insight article: https://www.physicsforums.com/insights/partial-differentiation-without-tears/