How do you differentiate \[y = \arcsin \left( x \right)\]?

Hint: Here we will differentiate the given function with respect to \[x\]. We will use the fact that \[arc\] is used to denote inverse a function, using this we will write the value in terms of \[x\]. Then we will use the differentiation method and the Pythagorean identity to get the desired answer.

Complete step-by-step answer:
We have to differentiate \[y = \arcsin \left( x \right)\] with respect to \[x\].
So, we can rewrite the value as,
\[y = {\sin ^{ - 1}}x\]
\[x = \sin y\]……\[\left( 1 \right)\]
Now differentiating both side with respect to \[x\] we get,
\[ \Rightarrow \dfrac{{d\left( x \right)}}{{dx}} = \dfrac{{d\left( {\sin y} \right)}}{{dx}}\]…..\[\left( 2 \right)\]
Using the differentiation formula \[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\], we get
\[\begin{array}{l} \Rightarrow 1 \times {x^{1 - 1}} = \cos y \times \dfrac{{dy}}{{dx}}\\ \Rightarrow 1 = \cos y \times \dfrac{{dy}}{{dx}}\end{array}\]
On cross multiplication, we get
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos y}}\]
Now using Pythagorean identity \[{\sin ^2}y + {\cos ^2}y = 1\], we can write
\[\cos y = \pm \sqrt {1 - {{\sin }^2}y} \]
Using this in above equation, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}y} }}\]
Using equation \[\left( 1 \right)\] in above value, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}\]
Therefore, we get differentiation of \[y = \arcsin \left( x \right)\] as \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\].

Note: Differentiation is used to calculate the instantaneous rate of change in the function given because of one of its variables. Differentiation is done with respect to an independent variable of the function. Some real life applications of differentiation is rate of change of velocity with respect to time. It is also used to find the tangent and normal curve as also to calculate the highest and lowest point of the curve in a graph. Differentiation of trigonometric function is a very vast topic where differentiation of different trigonometric values has different formulas. One important point that can be noted is we took the positive root from the Pythagorean identity because it is only possible for\[y = \sin x\] to have an inverse if we restrict the domain to: \[ - \dfrac{\pi }{2} \le x \le \dfrac{\pi }{2}\].