What is the derivative of y=arcsin⁡(x)?

Question

Anzante2m · Accepted Answer

This identity can be proven easily by applying

\sin

to both sides of the original equation:

1) y = \arcsin x

2) \sin y = \sin (\arcsin x)

3) \sin y = x

We continue by using implicit differentiation, keeping in mind to use the chain rule on

\sin y

4) \cos y \frac{d y}{d x} = 1

Solve for \frac{d y}{d x}

5) \frac{d y}{d x} = \frac{1}{\cos y}

Now, substitution with our original equation yields

\frac{d y}{d x}

in terms of

x

6) \frac{d y}{d x} = \frac{1}{\cos (\arcsin x)}

At first this might not look all that great, but it can be simplified if one recalls the identity

\sin (\arccos x) = \cos (\arcsin x) = \sqrt{1 - x^{2}}

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

This is a good definition to memorize, along with

\frac{d}{d x} [\arccos x] and \frac{d}{d x} [\arctan x]

since they appear quite frequently in differentiation problems.

SlabydouluS62 · Accepted Answer

Explanation:There is another way to do it. We have a formula for calculating the derivatives of inverse functions:(f−1)′(x)=1f′(f−1(x))Here f−1(x)=arcsin⁡x⇒f(x)=sin⁡x  and  f′(x)=cos⁡xSo (arcsin⁡x)′=1cos⁡(arcsin⁡x)Using the arguments given in the other answers, we know that1cos⁡(arcsin⁡x)=11−x2

RizerMix · Accepted Answer

Are you a professor of mathematics? He described everything very accurately and clearly explained

What is the derivative of y=\arcsin(x)?

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