Confusion in chain rule of differentiation

Question

Let sin(wt ) be the one.

Let z = wt

Then , we find dz / dt = w .

Then , let y = sin z

I am having confusion to why dy/dz will be - cos z and not only - cos .

Since , in dz /dt , t gets cut right . Why not in dy/dz also ?

Answer 1

So, there are several confusions. To begin with, there is no such thing as just $\cos$. It alway requires an argument. Also, I am unclear of whether $w$ is supposed to be a variable or a constant. Let's start with $w$ as a variable.

So, with $z = wt$, and $y = \sin(z)$, then

$$ \frac{dy}{dz} = \cos(z)\frac{dz}{dz} \\ \frac{dy}{dz} = \cos(z)$$

Note that we take the derivative of the interior and multiply the result by that derivative. It turns out to be $1$ ($\frac{dz}{dz}$), but I put in the step to be clear.

Now, let's find $\frac{dz}{dt}$ for $z = wt$. If $w$ is a variable, then, because of the product rule:

$$ \frac{dz}{dt} = w\frac{dt}{dt} + t\frac{dw}{dt}$$

But the $\frac{dt}{dt}$ simplifies to $1$, giving:

$$ \frac{dz}{dt} = w + t\frac{dw}{dt}$$

Now, if $w$ is a constant, then $w$ is not changing, and therefore $dw$ (the change in $w$) is zero. So, if $w$ is a constant, this simplifies to:

$$ \frac{dz}{dt} = w + 0 = w $$

So, since $\frac{dy}{dt} = \frac{dy}{dz}\frac{dz}{dt}$, then:

$$ \frac{dy}{dt} = \frac{dy}{dz}\frac{dz}{dt} = \cos(z)\,w \\ \frac{dy}{dt} = \cos(wt)\,w $$

Or, as they probably want you to write it, $$\frac{dy}{dt} = w\,\cos(wt)$$