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:

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

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:

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

And this implies:

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

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

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

Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:

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