# Chain rule

1. Jan 23, 2016

### Joseph1739

Not sure if this is the correct place to post this.

dy/dt = 0, find y(t)

My professor told me that the chain rule is used to determine that (dy/dt)*dt = dy, but I just don't see it.

Multiply both sides by dt.
(dy/dt)*dt = 0dt
(dy/dt)*dt = 0
dy = 0, then integrating both sides:
y = C

dy/dt is the derivative of the function y in respect to t, and dt is just a small change in t. How is the chain rule used here? dz/dx = dz/dy * dy/dx makes a lot more sense to me, but I just don't see that here.

2. Jan 23, 2016

### Dr. Courtney

I don't tend to explain that one with the chain rule or with the fundamental theorem of Calculus, but rather with the definition of derivative.

If dy/dt = 0, y(t) is a function whose derivative with respect to time is zero. Since derivative is the time rate of change, the derivative can only be zero (for all values of t) if the function y(t) is constant in time.

The only function whose derivative is zero at all input values is a constant.

3. Jan 24, 2016

### Joseph1739

That makes sense. Is the chain rule even applicable here though? dy/dt is not a fraction, so I don't see how you can multiply by dt then cancel the dt's. Then I don't believe dt is a derivative (isn't it a small change in t?), so how can the chain rule be used here? I don't want to struggle to do more complex problems in the future because I don't understand why dy/dt * dt = dy. I was able to find:
https://en.wikipedia.org/wiki/Differential_(infinitesimal)
but I still don't understand how the dx's cancel.

4. Jan 24, 2016

### Dr. Courtney

You can think of dy/dt as the ratio of differentials dy and dt. Recall the limit definition of derivatives. Consider the numerator to be dy and the denominator to be dt. This allows things like:

dy/dt = -k*y (Given)

dy = -k*y dt (Multiply both sides by dt)

1/y dy = -k dt (divide both sides by y)

Integrate both sides to get

ln(y) = -k*t + C

Etc.

5. Feb 1, 2016

### Staff: Mentor

Often this works. Sometimes it does not.
"Multiply with dt" and similar are ugly transformations that can lead to the correct result, but mathematically those operations are not well-defined.

Let y=2x.
dy/dt=0
Multiply by dt? dy=0.
Divide by dx? dy/dx=0.
Oops.

6. Feb 1, 2016

### Staff: Mentor

I'm not following here. If y = 2x, then the differentials involved would be dy and dx, not dy and dt.
$y = 2x$
$\Rightarrow \frac{dy}{dx} = 2$
Multiply by dx:
$dy = 2dx$
Am I missing something?

7. Feb 1, 2016

### Staff: Mentor

What stops me to make a derivative with respect to a third variable t?
Write it as y=2*x + 0*t if you like.

8. Feb 1, 2016

### MrAnchovy

Then you can only write $\frac{\partial y}{\partial t} = 0$ and we have $\frac{dy}{dx} = \frac{\partial y}{\partial x} + \frac{\partial y}{\partial t} \frac{dt}{dx} = 2 + 0 \cdot 0$. I do agree with your fundamental point though: it is dangerous to treat $dx$ as an independent entity that you can multiply and divide by as if it were a real number unless you have a firm grasp of the underlying analysis so that you know when that short cut cannot be used.

9. Feb 1, 2016

### Staff: Mentor

In post #5 you wrote this:
From this I infer that y is a function of an independent variable x. There was no mention of y also being a function of t. So without that additional information, taking the derivative with respect to t doesn't make any sense.