$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle

Question

The question contains a hint: at the appropriate point use the result $a^{3} - b^{3} = \left(a - b\right)\left(a^{2} + b^{2} + ab\right)$. $$ \frac{{\rm d}}{{\rm d}x}\sin\left(x^{1/3}\,\right) $$

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My attempt:

I did not use the hint as it was not immediately obvious to me what the simplification was. Instead I wrote with binomial expansion that $(x+h)^{\frac{1}{3}}=x^{\frac{1}{3}}(1+\frac{h}{x})^{\frac{1}{3}} \approx x^{\frac{1}{3}}(1+(\frac{1}{3})\frac{h}{x})$, used $\sin(()+())=\sin()\cos()+\sin()\cos()$ and applied $\cos() \approx 1$, $\sin() \approx ()$ for small $h$ which gave the correct answer.

But I do not see the point of using the cubes difference formula given in the question? Could this be a typo because it would make more sense to apply the identity to something like e.g. $\sin({x})^{\frac{1}{3}} (\text{not }\sin(x^{\frac{1}{3}}))\to \sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}} \iff (\sin(x+h)-\sin(x)=(\sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}})(\sin(x+h)^2+\sin(x)^2+\sin(x+h)\sin(x))$

Or am I missing something here?

Answer 1

You have

$$\begin{aligned} \frac{\sin(x^{\frac{1}{3}}) - \sin(c^{\frac{1}{3}}) }{x-c} &= \frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{x-c}\\ &=\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{(x^{\frac{1}{3}} - c^{\frac{1}{3}})(x^{\frac{2}{3}} + c^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}})} \end{aligned}$$

Now as $$\lim\limits_{x \to c }\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)}{x^{\frac{1}{3}} - c^{\frac{1}{3}}} = 1$$

we get $$\frac{d}{dx}(\sin(x^{\frac{1}{3}})) = \frac{\cos(x^{\frac{1}{3}})}{3 x^{\frac{2}{3}}}$$

Answer 2

hint

Replace in

$$\frac{a-b}{a^3-b^3}=\frac{1}{a^2+ab+b^2}$$

$ a \;\; \text{by} \;\; (x+h)^\frac 13 $

and

$ b\;\; \text{by}\;\; x^\frac 13$.