Multiplication of original function and derivative

Question

I am trying to simplify and calculate the following equation

$\frac{d y(t)}{dt}y(t)$

and I thought I was allowed to write it as

$\frac{d}{dt}y(t)\cdot y(t) = \frac{d y^2(t)}{dt}$

However the result of the calculations aren't the same or at the least confusing.

Using the chain rule I get

$\frac{d y^2(t)}{dt} = 2\cdot \ y(t) \cdot \frac{d y(t)}{dt} = 2 \cdot \frac{d y^2(t)}{dt} $

Which is the original function multiplied by two.

On a similar note, I also figured that

$\frac{d y(t)}{dt}\cdot \frac{1}{y(t)}=\frac{d }{dt}\cdot \frac{y(t)}{y(t)}=\frac{d}{dt} 1 = 0 $

Is this correct? If not, what am I doing wrong?

Answer 1

In general, we have $\frac{df}{dt}g \neq \frac{d(fg)}{dt}$

In fact, product rule says that $\frac{d(fg)}{dt}=\frac{df}{dt}g\color{blue}{+f\frac{dg}{dt}}$

Letting $f=g=y$,

we have $$\frac{dy^2}{dt}=2y\frac{dy}{dt}$$

$$\frac12\frac{dy^2}{dt}=y\frac{dy}{dt}$$

Also, $$\frac{d y}{dt}\frac{1}{y}=\frac{d(\ln y)}{dt}$$

Answer 2

As you noted, it follows from the chain rule that the derivative of $y^2$ is $2yy'$. What made you think that it was $yy'$? Or you can apply the rule of derivation of products but you will get, again, that $(y^2)'=2yy'$.

And, no $y'.\frac1y$ is not equal to $0$ in general. It's equal to $\frac{y'}y$.