When we take this derivative, for example:
$$y = \log(\sin x)$$ We call $u = \sin x$, so we have:
$$\frac{dy}{dx} = \frac{d y}{du}\frac{du}{dx} = \frac{1}{u}\cos x = \frac{\cos x}{\sin x}$$
But for me, it's better to do:
$$\frac{d\log\color{Blue}{\sin x}}{d\color{Blue}{\sin x}}\frac{d\sin \color{Red}{x}}{d\color{Red}{x}} = \frac{1}{\color{Blue}{\sin x}}\cos \color{Red}{x}$$ It makes easy to do the 'pattern-matching' just by looking at the differentials. No substitution. I know that $\frac{d \log[\mbox{something}]}{d[\mbox{something}]} = \frac{1}{\mbox{something}}$ for example.
However, it looks 'hairy' when I try with larger derivatives, like, for the function:
$$(13x^2-5x+8)^{\frac{1}{2}}$$ we do: $$\frac{d(13x^2-5x+8)^{\frac{1}{2}}}{dx} = \frac{d\color{Green}{(13x^2-5x+8)}^{\frac{1}{2}}}{d\color{Green}{(13x^2-5x+8)}}\frac{d(13x^2-5x+8)}{dx} = \frac{1}{2\sqrt{\color{Green}{13x^2-5x+8}}}(26x -5)$$ but it's really better for me to do like this, instead of doing the bla bla bla of changing variables and stuff. But I'm afraid my teacher does not accept this. Is this notation/way of doing good for you guys?
One more example
: $$\frac{d}{dx}\sqrt{(\sin(7x+\ln(5x)))} = $$
$$\frac{d[\color{Blue}{\sin(7x+\ln(5x))}]^{1/2}}{d[\color{Blue}{\sin(7x+\ln(5x))}]}\frac{d[\sin\color{Red}{(7x+\ln(5x))}]}{d[\color{Red}{7x+\ln(5x)}]}\left[\frac{d[7\color{Purple}{x}]}{d[\color{Purple}{x}]} + \frac{d[\ln(\color{Purple}{5x})]}{d[\color{Purple}{5x}]}\frac{d[5x]}{d[x]}\right] = $$ $$\frac{1}{2}\left[\color{Blue}{\sin(7x+\ln(5x))}\right]^{-1/2}\cdot\cos(\color{Red}{7x+\ln(5x)})\left[7 + \frac{1}{\color{Purple}{5x}}\cdot 5\right]$$ So we get rid of the substitution!
(づ。◕‿‿◕。)づ $\ \ u, v, y$ go away!