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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.

Vincenzo Tibullo
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Koh Yi Min Jason
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  • Well the equation u=sin x is solvable for x, and lets you write x as a function of u: what you want to do is then simply the chain rule. – Mariano Suárez-Álvarez Apr 23 '23 at 10:06
  • Ahhh, thanks for your hint. so the new equation with u = sin(3x) would be $f(\frac{arcsin(u)}{3})= u^{4} + \frac{4}{9}(arcsin(u))^{2} + 3g(\frac{arcsin(u)}{3})$ – Koh Yi Min Jason Apr 23 '23 at 10:17
  • Does this answer your question? Differentiating with respect to a function – Anne Bauval Apr 23 '23 at 10:23
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    To be honest, the notation is bad. You need to define exactly what you mean by $df/dh$ for a function $h$. – durianice Apr 23 '23 at 10:30
  • @AnneBauval No, that is for muti-variable differentiation of a function. I am looking at single variable differentiation of a function rather than a variable as such, I do not want to be confused with many other equations. – Koh Yi Min Jason Apr 23 '23 at 10:31
  • If we define $h_1(x,y,z)=y^4+4x^2+3z$ and $h_2(x)=(x,\sin(3x),g(x))$, and we define "$df/d\sin(3x)$" to mean $\frac{\partial h_1}{\partial y}(h_2(x))$, then $\frac{df}{d\sin(3x)}(x)=4\sin(3x)^3$. – durianice Apr 23 '23 at 10:32
  • @AnneBauval Anyway, thanks for sharing that question. I have already figure out how to solved the problem. – Koh Yi Min Jason Apr 23 '23 at 10:32
  • If we define $h_1(x)=x^4+4(\arcsin(x)/3)^2+3g(\arcsin(x)/3)$ and $h_2(x)=\sin(3x)$, and we define "$\frac{df}{d\sin(3x)}$" to mean $\frac{d(h_1\circ h_2)}{dx}$, then the $u$-substitution works. – durianice Apr 23 '23 at 10:34
  • I’m voting to close this question because the answer is straightforward if you only decide if you need a partial or a total derivative of $f$ w.r.t. $\sin(3x),.$ Simple plain calculus in the end. – Kurt G. Apr 23 '23 at 15:35

3 Answers3

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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

UnsoundMouse
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    Your answer ignores the possibility that OP did not clearly mention if a total or a partial derivative is required. The entire post is of very low quality and should be closed. – Kurt G. Apr 23 '23 at 15:34
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Assuming that $f(x)=\sin^4(3x)+4x^2+3g(x)$, I would solve by differentials like that:

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

Bob Dobbs
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Hint

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

MathStackexchangeIsNotSoBad
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