Inverse Trig Functions Revision Notes for Leaving Cert Mathematics

Inverse Trig Functions

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions are useful in calculus, particularly in integration and solving implicit differentiation problems.

The key differentiation rules are as follows:

\frac{d}{dx} \cos^{-1} \left(\frac{x}{a}\right) = \frac{-1}{\sqrt{a^2 - x^2}}

\frac{d}{dx} \sin^{-1} \left(\frac{x}{a}\right) = \frac{1}{\sqrt{a^2 - x^2}}

\frac{d}{dx} \tan^{-1} \left(\frac{x}{a}\right) = \frac{a}{\sqrt{a^2 + x^2}}

These formulas are particularly useful when differentiating functions that involve inverse trigonometric expressions.

In many cases, the chain rule is required when differentiating more complex expressions involving inverse trigonometric functions.

Example

infoNote

Differentiate the function $y=\sin^{-1} \left( \frac{1}{x^2}\right)$ with respect to $x$ .

Identify an inner and outer function :

u=\sin^{-1}v=\sin^{-1}\left(\frac{v}{1} \right)

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

Differentiate :

u'=\frac{1}{\sqrt{1-v^2}}

v'=-\frac{2}{x^3}

Apply chain rule :

u' \cdot v'= \frac{1}{\sqrt{1-v^2}} \cdot -\frac{2}{x^3}

= -\frac{2}{x^3\sqrt{1-v^2}}

= -\frac{2}{x^3\sqrt{1-\left(\frac{1}{x^2} \right)^2}}

= -\frac{2}{x^3\sqrt{\left(\frac{x^4-1}{x^4} \right)}}

= -\frac{2}{x\sqrt{x^4-1}}

Inverse Trig Functions (Leaving Cert Mathematics): Revision Notes