# Is this notation good for the chain rule derivative?

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!

• This is probably the coolest (most decorated) question asked. (づ｡◕‿‿◕｡)づ – Shahar May 16 '14 at 23:16
• @Shahar thank you (｡◕‿◕｡) – Lucas Zanella May 16 '14 at 23:18
• @LucasZanella I must say, I'm really looking forward to this question's answer. I'm a self-learner, and do not know much about the commonly used notation. – user122283 May 16 '14 at 23:20
• As for the question, it's right but not very conventional. Hence, I'd assume your teacher might not accept it (depends on how he taught it). But you're doing it right. (｡◕‿◕｡) – Shahar May 16 '14 at 23:20
• omg you guys are so nice, I love this forum ♥ – Lucas Zanella May 16 '14 at 23:21

Be warned that this notation does get overcomplicated at times and you will want to add some substitutions for sanity, but for simpler calculations, it can be useful and looked upon favorably. You should look to study differential calculus as a topic in its own right, where this notation is used as well (in summary, the "denominator" of the derivative $$\text{d}[\text{something such as} \,x]$$ is eliminated and we just deal with what are called differentials). Your instructor is very intolerant, indeed, if he doesn't allow this.