MHB JustWar's question at Yahoo Answers regarding differentiating a power of arcsin

[MarkFL]

Here is the question:

Find the derivative of arcsin^3(5x+5)?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello JustWar,

We are given to differentiate:

$$y=\arcsin^3(5x+5)$$

Once approach we may use is if we don't happen to know the formula for differentiating the inverse sine function to take the cube root of both sides:

$$y^{\frac{1}{3}}=\arcsin(5(x+1))$$

And this implies:

$$5(x+1)=\sin\left(y^{\frac{1}{3}} \right)$$

Now, implicitly differentiating with respect to $x$, we obtain:

$$5=\cos\left(y^{\frac{1}{3}} \right)\left(\frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx} \right)$$

Solving for $$\frac{dy}{dx}$$, we obtain:

$$\frac{dy}{dx}=15y^{\frac{2}{3}}\sec\left(y^{\frac{1}{3}} \right)=\frac{15\arcsin^2(5(x+1))}{\sqrt{1-(5(x+1))^2}}$$