Partial derivative of $\sin(xz)$: Is this making sense and is it a proper use of notation?

Question

The exercise here is to calculate $\dfrac{\partial}{\partial x} \sin(xz)$

The usual way is just considering $z$ as a constant and deriving, so it would be $z\cos(xz)$.

However, I want to solve it this way: Define $a=xz$, then $\sin(a) = \sin(xz)$

$\dfrac{\partial}{\partial x} \sin(a) = \dfrac{\partial}{\partial a} \sin(a) \dfrac{\partial}{\partial x} xz = z\cos(a) = z\cos(xz)$.

Is this second way correct? My friend said it is not making sense and not a proper use of notation. But I do not understant why...

Answer 1

(Just to not leave this unanswered)

As comments suggested, this is a fair use of chain rule and there is nothing wrong. The other given answers are talking about problems that doesn't exist.

Answer 2

Yes.. but you are trying to replicate the chain rule in derivatives with the problem being it's not applied correctly. You can not use /∂a for Sin(a) and /∂x for xz.

You'll still get the same answer if you correct your technique slightly -

Here you go -

The Chain Rule says: derivative of f(g(x)) = f'(g(x))g'(x)

where

f(g) = sin(g)

Considering a=xz
Taking Sine on both sides sin(a) = sin(xz)

g(x) = xz

∂/∂x sin(a)=∂/∂x (Sin (xz))

-- Applying Chain Rule in Derivatives

0 = Cos(xz).∂/∂x(xz)

=zCos(xz)