# Chain rule

1. Aug 8, 2017

### Karol

1. The problem statement, all variables and given/known data

Isn't the derivative of y with respect to x Defined as $~\frac{dy}{dx}$?
What and how do i have to prove?

2. Relevant equations
The chain rule:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

3. The attempt at a solution
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

2. Aug 8, 2017

### Staff: Mentor

What a horrible question. dy and dx are not numbers, and dx/dy is not a ratio.

What you can probably do is studying d y(x(t))/dt and then abusing the notation sufficiently until you get dy/dx, showing that t does not matter.

3. Aug 8, 2017

### Karol

What is y[x(t)]? isn't it just y(t)?
For example, the example given in the book:
$$x=t+\frac{1}{t},~~y=t-\frac{1}{t}~\rightarrow~x^2-y^2=4$$
isn't $~y[x(t)]=t-\frac{1}{t}~$?

4. Aug 8, 2017

### Staff: Mentor

No. What is $y(x)$ if $x$ was just a real number?

5. Aug 8, 2017

### Ray Vickson

In Nonstandard Analysis, $dy$ and $dx$ are actual infinitesimals and $dy/dx$ really is a ratio; the "standard part" of the ratio equals the ordinary derivative.

However, I have serious doubts that the OP's course is using full-fledged non-standard analysis, but is instead likely using a form of heuristic argument that has been around since the days of Leibnitz.

6. Aug 9, 2017

### Karol

$$y=\sqrt{(x-2)(x+2)}$$

7. Aug 9, 2017

### Staff: Mentor

This is not how $y$ is defined.

8. Aug 9, 2017

### Karol

If x were a real number then $~y=x-\frac{1}{x}$
But what is y[x(t)]?
$$y[x(t)]=y\left[ t+\frac{1}{t} \right]=\left(t+\frac{1}{t} \right)-\frac{1}{\left(t+\frac{1}{t} \right)}\neq t-\frac{1}{t}$$
I don't understand

Last edited: Aug 9, 2017
9. Aug 9, 2017

### Staff: Mentor

You have $y(sth.)= sth. \,- \frac{1}{sth.}$. Next this something is $x(t) = sth. = t + \frac{1}{t}$. What do we get by substitution of that something?

10. Aug 9, 2017

### Karol

We get the value of y, but y isn't a function of x, each (x and y) are functions of t. we just join the points (x,y) produced by t.

11. Aug 9, 2017

### Staff: Mentor

You have $y(t)=t-\frac{1}{t}$. This means, if $t$ is replaced by $x$, then we get $y(x)=x-\frac{1}{x}$. And if then the function $x=x(t)= t + \frac{1}{t}$ is parameterized by $t$, we get
$$(y \circ x)(t)=y(x(t))=x(t) -\frac{1}{x(t)} = (t+\frac{1}{t})- \frac{1}{t+\frac{1}{t}}$$
Now you can calculate either $\frac{d}{dt}(y\circ x)(t)$ as function of $t$ or use the chain rule and calculate
$$\frac{d}{dt}(y\circ x)(t)= \frac{d}{dx}y(x) \cdot \frac{d}{dt}x(t) = \frac{d}{dx} (x-\frac{1}{x}) \cdot \frac{d}{dt} (t+\frac{1}{t})$$
and then substitute $x=t+\frac{1}{t}$ and hope both quotients are equal.

12. Aug 10, 2017

### Karol

It is very tiring to prove that both these equations are equal.
And even so, what did i prove? that $~\frac{d}{dx}y(x) \cdot \frac{d}{dt}x(t) = \frac{d}{dt}y[x(t)]$
Is it what the question asked? in this example:
$$y=\sqrt{(x-2)(x+2)}~\rightarrow~y'(x)=...$$
And i have to prove it equals dy/dx
I leave this question since i don't want to tire you and me, i think i won't get further.
Thank you fresh_42

13. Aug 10, 2017

### Ray Vickson

No: there are two solutions
$$y(x) = \pm \sqrt{(x-2)(x+2)}$$
This really is true: if you plot the points $(x(t),y(t))$ for $t> 0$ you will find they lie on the curve $y = -\sqrt{(x-2)(x+2)}$ for $0 < t \leq 1$ and on the curve $y = + \sqrt{(x-2)(x+2)}$ for $1 \leq t < \infty$. This is one branch of the hyperbola $x^2 - y^2 = 4$. For $t < 0$ you get the opposite branch of the hyperbola, opening out to the left of $x = -2$.

14. Aug 10, 2017

### Karol

Thank you Ray, you all are specialists, they drew the plot in the book and it's as you've said