What is the derivative of $arcsin (x^{\frac{1}{2}})$ $arcsin(x^(1/2))$ ?

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What is the derivative of $arcsin (x^{\frac{1}{2}})$ $arcsin(x^(1/2))$ ?

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Answer 1 · 2016-06-30T17:39:35

$\frac{d y}{d x} = \frac{1}{2 (\sqrt{x (1 - x)})}$ $(dy)/(dx) = 1/(2(sqrt(x(1-x))))$

Explanation:

First I'm going to walk you through a nice way of deriving the inverse trig derivatives.

Start with $y = arcsin (x)$ $y = arcsin(x)$ ,this allows us to construct a triangle shown below. enter image source here

Rearranging we get that $x = sin (y)$ $x = sin(y)$

Hence, $\frac{d x}{d y} = cos (y)$ $(dx)/(dy) = cos(y)$

Note that $cos (y) = \sqrt{1 - x^{2}}$ $cos(y) = sqrt(1-x^2)$

We obtain that:

$\frac{d y}{d x} = \frac{1}{cos (y)} = \frac{1}{\sqrt{1 - x^{2}}}$ $(dy)/(dx) = 1/(cos(y)) = 1/(sqrt(1-x^2))$

Now that we know the general form, we use the chain rule.

For $y (u (x)), \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $y(u(x)), (dy)/(dx) = (dy)/(du)*(du)/(dx)$

So $\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(x^{\frac{1}{2}})}^{2}}} \cdot \frac{d}{d x} (x^{\frac{1}{2}})$ $(dy)/(dx) = 1/(sqrt(1-(x^(1/2))^2))*d/(dx)(x^(1/2))$

$\frac{d y}{d x} = \frac{1}{2 (\sqrt{x (1 - x)})}$ $(dy)/(dx) = 1/(2(sqrt(x(1-x))))$

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What is the derivative of $arcsin (x^{\frac{1}{2}})$ $arcsin(x^(1/2))$ ?

1 Answer

Explanation:

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What is the derivative of $arcsin (x^{\frac{1}{2}})$ $arcsin(x^(1/2))$ ?