Question

What are the derivatives of the inverse trigonometric functions?

Answer 1

See below.

Explanation:

`d/dxsin^-1x=1/sqrt(1-x^2)`

`d/dxcos^-1x=-1/sqrt(1-x^2)`

`tan^-1x=1/(1+x^2)`

`cot^-1x=-1/sqrt(1+x^2)`

`sec^-1x=1/(xsqrt(x^2-1))`

`csc^-1x=-1/(xsqrt(x^2-1))`

One useful thing to notice is that the derivatives of all inverse "co" functions are equivalent to the derivatives of the original inverse function, but have a negative added.

Here's a proof for the derivative of the inverse sine function, if you want to avoid memorization. All the other ones can be proved in the same way.

`y=sin^-1x hArr x=siny`, from the definition of an inverse function.

Differentiating `x=siny:`

`d/dx(x)=d/dx(siny)`

`1=cosy*dy/dx` (Implicit Differentiation)

`dy/dx=1/cosy`

We need to get rid of `cosy.` Recall that we said `x=siny` and recall the identity `sin^2theta+cos^2theta=1`. This can be rewritten for `y` and solved for cosine as follows:

`cos^2y=1-sin^2y`

`cosy=sqrt(1-sin^2y)`

Recalling that `x=siny,` then `sin^2y=x^2`

Thus,

`cosy=sqrt(1-x^2)`

`d/dxsin^-1x=1/sqrt(1-x^2)`

I recommend that you commit these integrals to memory- they will help you when you are learning the trig. substitution.