Inverse Trig Derivatives

This is a nice trick for calculating the derivatives of inverse trig functions.  I’ll demonstrate the process for the  derivative of \sin^{-1}\theta.  The rest of the inverse trig derivatives can be found using the same method.

First, recall the definitions of the trig functions on the unit circle (Figure 1 below)

trig-func-der1

Figure 1

The inverse sine function is defined by the property

\displaystyle \sin^{-1}(\sin\theta)=\theta

If we define

\displaystyle x=\sin \theta

and re-lable the diagram in terms of x (Figure 2)

trig-func-der2

Figure 2

Now define y

\displaystyle y=\sin^{-1} (x)

then invert the equation

\displaystyle x= \sin y

Differentiate the equation with respect to x

\displaystyle 1= \cos y \frac{dy}{dx}

Then

\displaystyle \frac{dy}{dx}= \frac{1}{\cos y}

From the definition of y and Figure 2, we see

\displaystyle \cos y=\cos[\sin^{-1}(x)]=\sqrt{1-x^2}

Then

\displaystyle \frac{dy}{dx}= \frac{d}{dx}\sin^{-1}(x)= \frac{1}{\sqrt{1-x^2}}

and thus we have found the derivative of the inverse sine!  The same method can be used to find all of the other inverse trig derivatives.  You have to think about what should be labeled x, but it is rather straightforward.  It is then possible to construct a table of inverse trig derivatives, which can be used as an integral table. For instance

\displaystyle \int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)

which can be generalized to

\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(x/a)

where a is a constant.