What is the derivative of arcsin [ x 1 2 ] ?

Question

What is the derivative of             arcsin              [                  x                                    1              2                                      ]            ?

anajusthings5mrf · Accepted Answer

To find the derivative we will need to use the Chain Rule

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

We want to find

\frac{d}{d x} (\arcsin (x^{\frac{1}{2}}))

Following the chain rule we let

u = x^{\frac{1}{2}}

Deriving u we get

\frac{d u}{d x} = \frac{1}{2} \cdot x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}

Then, we substitute u in place of x in the original equation and derive to find

\frac{d y}{d u}

y = \arcsin (u)

\frac{d y}{d u} = \frac{1}{\sqrt{1 - u^{2}}}

Next, we substitute these derived values into the chain rule to
find

\frac{d y}{d x}

\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}

\frac{d y}{d x} = \frac{1}{\sqrt{1 - u^{2}}} \cdot \frac{1}{2 \sqrt{x}}

Substitute x back into the equation to get the derivative in terms of x only and simplify

u = x^{\frac{1}{2}}

\frac{d y}{d x} = \frac{1}{\sqrt{1 - {(x^{\frac{1}{2}})}^{2}}} \cdot \frac{1}{2 \sqrt{x}}

\frac{d y}{d x} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2 \sqrt{x}}

\frac{d y}{d x} = \frac{1}{2 \sqrt{x} \cdot \sqrt{1 - x}}

\frac{d y}{d x} = \frac{1}{2 \sqrt{x - x^{2}}}

What is the derivative of arcsin[x^(1/2)]?

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