# I partial derivatives

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 ?