0
$\begingroup$

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:

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

$\endgroup$

1 Answer 1

1
$\begingroup$

They are both correct and they are equal to each other, but ${\sqrt {1 - {x^2}} }$ is much easier to compute and read than ${\cos \left( {{{\sin }^{ - 1}}x} \right)}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.