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

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To differentiate the function y = arcsin^3(5x + 5), an effective method involves taking the cube root of both sides. This leads to the equation 5(x + 1) = sin(y^(1/3)), which is then implicitly differentiated with respect to x. The resulting expression for the derivative is dy/dx = 15y^(2/3)sec(y^(1/3)). Substituting back, this can be expressed as dy/dx = 15arcsin^2(5(x + 1))/sqrt(1 - (5(x + 1))^2). This approach effectively illustrates the differentiation of the arcsine function raised to a power.
MarkFL
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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.
 
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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:

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