# 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!

• One problem: $Q$ is a function of two variables, not one. – Randall May 20 '19 at 1:50
• $2x$ and $\sqrt[3]{x+1}$ are multiplied together, and require the use of the product rule inside the chain rule. – The Count May 20 '19 at 1:54
• @Randall then what should I do? – AaronTBM May 20 '19 at 2:00
• @TheCount yess, I did so – AaronTBM May 20 '19 at 2:01
• Do it in two pieces, not three. – Randall May 20 '19 at 2:03

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

Hint:

$$\bigg(\sin[f(x)]\bigg)'=f'(x)\cos[f(x)]$$

by the chain rule.

Then by the product rule: $$(2x\cdot(x+1)^\frac13)'=(2x)((x+1)^\frac13)'+(2x)'(x+1)^\frac13$$

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 $$\cos(2x. \sqrt[3]{x+1}).[2\sqrt[3]{x+1} + (\frac{2x}{3\sqrt[3]{(x+1)^2}}).2(x+1)]$$