trigonometry - Differentiating the function $arcsin(3x-4x^3)$

When I have to differentiate the function $\arcsin(3x-4x^3)$ which of the following methods is more appropriate ?

1. Putting $x=\sin θ$,simplifying and then differentiating for certain ranges of $x$.



2. Directly differentiating using chain rule.

Can the results obtained in these two techniques be shown to be same?
BTW I really don't understand why most textbooks prefer the first method. Any ideas? Thank you.
P.S:I know how to differentiate it.My question is something else ^ .

Answer

$$1-(3x-4x^3)^2=1-9x^2+24x^4-16x^6$$

$$=1-x^2-8x^2(1-x^2)+16x^4(1-x^2)=(1-x^2)(1-8x^2+16x^4)$$

$$=(1-x^2)(1-4x^2)^2$$

Now $3-12x^2=3(1-4x^2)$

$\implies\dfrac{3-12x^2}{1-(3x-4x^3)^2}=\dfrac{3(1-4x^2)}{\sqrt{1-x^2}|1-4x^2|}$

Now $|1-4x^2|=+(1-4x^2)\iff1-4x^2\ge0\iff-\dfrac12\le x\le\dfrac12$

Again, $\arcsin(3x-4x^3)=3\arcsin x\iff-\dfrac\pi2\le3\arcsin x\le\dfrac\pi2$

$\iff-\dfrac\pi6\le\arcsin x\le\dfrac\pi6\iff-\sin\dfrac\pi6\le x\le\sin\dfrac\pi6$ i.e., $-\dfrac12\le x\le\dfrac12$

The rest I want to leave for you as an exercise