Derivative of Sine Inverse

In this tutorial we shall discuss the derivative of inverse trigonometric functions and first we shall prove the sine inverse trigonometric function.

Let the function of the form

y = f\left( x \right) = {\sin ^{ - 1}}x

By the definition of the inverse trigonometric function, y = {\sin ^{ - 1}}x can be written as

\sin y = x

Differentiating both sides with respect to the variable x, we have

\begin{gathered} \frac{d}{{dx}}\sin y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow \cos y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\cos y}}\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Since y is restricted in the interval \left] { - \frac{\pi }{2},\frac{\pi }{2}} \right[ for  - 1 < x < 1, \cos y can have only positive values, and from the fundamental trigonometric rules \cos y = \sqrt {1 - {{\sin }^2}y} . Putting this value in the above relation (i) and simplifying, we have

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\sin }^2}y} }} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }},\,\,\,\, - 1 < x < 1 \\ \Rightarrow \frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 - {x^2}} }},\,\,\,\, - 1 < x < 1 \\ \end{gathered}

 

Example: Find the derivative of

y = f\left( x \right) = {\sin ^{ - 1}}2x

We have the given function as

y = {\sin ^{ - 1}}2x

Differentiating with respect to variable x, we get

\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\sin ^{ - 1}}2x

Using the rule, \frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 - {x^2}} }}, we get

\begin{gathered} \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}\frac{d}{{dx}}2x \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{2}{{\sqrt {1 - 4{x^2}} }} \\ \end{gathered}