Chain rule in derivatives Well, I have to derive this function: $$f(x)=\sin(2x \sqrt[3]{x+1} )$$
I want to use the chain rule, and I want to use it like this;
I will call:
$$T=x+1$$
$$Q=2x. \sqrt[3]{t} $$
$$f=\sin (Q)$$
So then I have:
$$\cfrac{df}{dx} = \cfrac{df}{dQ} . \cfrac{dQ}{dT} . \cfrac{dT}{dx} . $$
And now I just have to derive. For example, 
$$\cfrac{df}{dQ} = \cos (Q) $$ and then I put $$2x. \sqrt[3]{t} $$ instead of Q, and so on.
But what should I do when I want to do $ \cfrac{dQ}{dT} $ ? Because I have a product, and I know that I have to use the product rule and it would be $$Q'=2x'.\sqrt[3]{t}+2x.(\sqrt[3]{t})'$$ but what shall I do when deriving that X in $(2x)'$? Because the derivative is $ \cfrac{dQ}{dT} $, not $ \cfrac{dQ}{dx} $
I know that I could avoid putting names to these "sub-functions", but this way is easier to me, so, please, don't teach me any other method.
Thanks!
 A: According to the chain rule, $(f(g(x)))'=g'(x)\cdot f'(g(x))$
So for the function, $f(x)=\sin(2x\sqrt[3]{x+1})$, will be differentiatd as follows...
By chain rule obviously, we get, $(2x\sqrt[3]{x+1})'\cos(2x\sqrt[3]{x+1})$...Then,
$$\text{Taking } (2x\sqrt[3]{x+1})' = (2x)'(\sqrt[3]{x+1})+(\sqrt[3]{x+1})'(2x)$$
$$= 2(\sqrt[3]{x+1})+\frac{2x}{3}(\sqrt{(x+1)^3})$$
Multipying this together with $\cos(2x\sqrt[3]{x+1})$,
$$\Biggl\{2(\sqrt[3]{x+1})+\frac{2x}{3}(\sqrt{(x+1)^3})\Biggr\}\cdot\cos(2x\sqrt[3]{x+1})$$
This is the derivative. Maybe converted into other forms
A: Hint:
$$\bigg(\sin[f(x)]\bigg)'=f'(x)\cos[f(x)]$$
by the chain rule.
Then by the product rule:
$$(2x\cdot(x+1)^\frac13)'=(2x)((x+1)^\frac13)'+(2x)'(x+1)^\frac13$$
A: As $$Q'=2x'.\sqrt[3]{t}+2x.(\sqrt[3]{t})'$$ you can write $x'$ as $$ x' = \frac{dx}{dT}$$
Then in the next step;$$2(\frac{dx}{dT}).(\frac{dT}{dx})\sqrt[3]{t} + (\frac{2x}{3\sqrt[3]{t^2}})(\frac{dT}{dx})$$
$$2.\sqrt[3]{x+1} + (\frac{2x}{3\sqrt[3]{(x+1)^2}}).2(x+1)$$
The full answer will be $$\cos(2x. \sqrt[3]{x+1}).[2\sqrt[3]{x+1} + (\frac{2x}{3\sqrt[3]{(x+1)^2}}).2(x+1)]$$
