I'm a bit confused about how the inverse trig functions are differentiated. From a website:
We have the following relationship between the inverse sine function and the sine function.
$$\sin \left( {{{\sin }^{ – 1}}x} \right) = x\hspace{0.5in}{\sin ^{ – 1}}\left( {\sin x} \right) = x$$
In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,
$$f\left( x \right) = \sin x\hspace{0.5in}g\left( x \right) = {\sin ^{ – 1}}x$$
Then,
$$g'\left( x \right) = \frac{1}{{f'\left( {g\left( x \right)} \right)}} = \frac{1}{{\cos \left( {{{\sin }^{ – 1}}x} \right)}}
$$This is not a very useful formula. Let’s see if we can get a better formula. [..]
Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $g'(x)$ is expressed in terms of $x$ which is what we want and we applied the inverse function rule correctly.
Is there a reason for continuing until we arrive at:
$$\frac{d}{{dx}}\left( {{{\sin }^{ – 1}}x} \right) = \frac{1}{{\sqrt {1 – {x^2}} }}
$$
Best Answer
They are both correct and they are equal to each other, but ${\sqrt {1 - {x^2}} }$ is much easier to compute and read than ${\cos \left( {{{\sin }^{ - 1}}x} \right)}$.