Derivative of Cosine Inverse

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

Let the function of the form be \[y = f\left( x \right) = {\cos ^{ – 1}}x\]

By the definition of inverse trigonometric function, $$y = {\cos ^{ – 1}}x$$ can be written as
\[\cos y = x\]

Differentiating both sides with respect to the variable $$x$$, we have
\[\begin{gathered} \frac{d}{{dx}}\cos y = \frac{d}{{dx}}\left( x \right) \\ \Rightarrow – \sin y\frac{{dy}}{{dx}} = 1 \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{\sin y}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\\ \end{gathered} \]

Since $$y$$ is restricted in the interval $$\left] {0,\pi } \right[$$ for $$ – 1 < x < 1$$, $$\sin y$$ can have only positive values, and from the fundamental trigonometric rules $$\sin y = \sqrt {1 – {{\cos }^2}y} $$. Putting this value in the above relation (i) and simplifying, we have
\[\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{1}{{\sqrt {1 – {x^2}} }},\,\,\,\, – 1 < x < 1 \\ \Rightarrow \frac{d}{{dx}}\left( {{{\cos }^{ – 1}}x} \right) = – \frac{1}{{\sqrt {1 – {x^2}} }},\,\,\,\, – 1 < x < 1 \\ \end{gathered} \]

Example: Find the derivative of \[y = f\left( x \right) = {\cos ^{ – 1}}{x^2}\]

We have the given function as
\[y = {\cos ^{ – 1}}{x^2}\]

Differentiating with respect to variable $$x$$, we get
\[\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\cos ^{ – 1}}{x^2}\]

Using the cosine inverse rule, $$\frac{d}{{dx}}\left( {{{\cos }^{ – 1}}x} \right) = – \frac{1}{{\sqrt {1 – {x^2}} }}$$, we get
\[\begin{gathered} \frac{{dy}}{{dx}} = – \frac{1}{{\sqrt {1 – {{\left( {{x^2}} \right)}^2}} }}\frac{d}{{dx}}{x^2} \\ \Rightarrow \frac{{dy}}{{dx}} = – \frac{{2x}}{{\sqrt {1 – {x^4}} }} \\ \end{gathered} \]