Differentiating with respect to an expression rather than a variable

Question

suppose if I have the following equation:

$$f(x) = \sin(3x)^{4} + 4x^{2} + 3g(x)$$

What is the derivative of $f(x)$ with respect to $\sin(3x)$?

Does $\frac{df(x)}{d\sin(3x)} = 4\sin(3x)^{3}$ even though the terms $4x^{2}$ and $3g(x)$ are functions of $x$?

Thanks.

Answer 1

Hint

$$\frac{d}{d\sin x}f(x)=\frac{\frac{d}{dx}f(x)}{\frac{d\sin x}{dx}}$$

Answer 2

i think that if you are going to derive with respect to something containing x then you can't treat anything containing that x as it is a funtion of it to derive this you just need to make a simple substitution

let: $\alpha=$$\sin(3x) $$\Leftrightarrow x=$$\frac13$$\arcsin($$\alpha) $

then the function becomes in terms of $\alpha$ as:
$f($$\alpha)=$$\alpha^4 + $4($\frac13$$\arcsin($$\alpha))² + $$3g($$\frac13$$\arcsin($$\alpha$))

which now you can derive easily with the power and chain rule to get:

$\frac{d}{d\alpha}f(\alpha)=4\alpha^3 + \frac{8}{9}\frac{\arcsin(\alpha)}{\sqrt{1-\alpha²}} + \frac{g'(\frac{1}{3}\arcsin(\alpha))}{\sqrt{1-\alpha²}}$

now you can try and substitute $\sin(3x)$ back in to get:

$\frac{d}{d\sin(3x)}f(x)=4\sin(3x)^3 + \frac{8}{9}(\frac{3x}{\sqrt{1-\sin²(3x)}}) + \frac{g'(x)}{\sqrt{1-\sin²(3x)}}$

you can also make it look a little better by replacing $\sqrt{1-\sin²(3x)}$ with $\cos(3x)$

and that's it

Answer 3

Assuming that $f(x)=\sin^4(3x)+4x^2+3g(x)$, I would solve by differentials like that:

$$df=(12\sin^3(3x)\cos(3x)+8x+3g'(x))dx$$ $$d\sin(3x)=3\cos(3x)dx$$ $$\begin{align} \frac{df}{d\sin(3x)}&=\frac{12\sin^3(3x)\cos(3x)+8x+3g'(x)}{3\cos(3x)}\frac{dx}{dx}\\ &=4\sin^3(3x)+\frac83x\sec(3x)+g'(x)\sec(3x) \end{align}$$