## Derivatives of inverse functions

Here, we introduce something that really should be introduced in

Now, think about the derivative of sin-1(x). Suppose y=sin-1(x). then we have sin(y)=x. Differentiating both sides using the chain rule, we have y'cos(y)=1. or, y'=1/cos(y).

The above example can be summarized as f'(x)=1/f'(y). This is true for all inverse functions.

For example, up there, we had y'=1/cos(y). Yet, y=sin-1(x). Now cos(sin-1(x))=sqrt(1-x^2).

Using the same logic, you can deduce some formulas for the rest of the derivatives of the inverse trigonometric functions, with no problem at all!

__the chain rule__section, but really, it kicks in here. (But knowledge of the chain rule is required, so I recommend giving the above link a click) Soon, we shall introduce the derivatives of inverse trigonometric functions. First, warm yourself up with inverse calculations: e.g. sin(cos-1(x))=sqrt(1-x^2).Now, think about the derivative of sin-1(x). Suppose y=sin-1(x). then we have sin(y)=x. Differentiating both sides using the chain rule, we have y'cos(y)=1. or, y'=1/cos(y).

*Note that we have used y' to denote f'(x). this is a quite common shorthand. If it is a shorthand at all.*The above example can be summarized as f'(x)=1/f'(y). This is true for all inverse functions.

**However, in trigonometry, we can further simplify f'(y) by obtaining a relationship between y and x, and substitute y for whatever of x.**For example, up there, we had y'=1/cos(y). Yet, y=sin-1(x). Now cos(sin-1(x))=sqrt(1-x^2).

__Correct me if i am wrong__. So now y'=1/sqrt(1-x^2).Using the same logic, you can deduce some formulas for the rest of the derivatives of the inverse trigonometric functions, with no problem at all!