## Get Arc Function Derivatives When You Forget Formulas

Have you ever come into a situations that you forget the derivative formulas for arcfunctions and cannot do the problems in exam? Well, that happens to a lot of people, and they lose points just because they don't remember the forumulas, and those formulas are not given in exams! But there's an easy way to solve this problem --- to find the formula yourself!

Try to find the formula for arcsin, arctan and arcsec first.Here is what you can do:

(arcsin x)' = ?

sin(arcsin x) = x  (since sin and arcsin are inverse functions)

[sin (arcsin x)]' = x'  (take derivatives on both sides)

Then we use chain rule and get:

cos(arcsin x) * (arcsinx)' = 1

(arcsinx)' = 1 / cos(arcsin x)

Now you have already get 70% done. And below we are going to use the trigonometry formulas to convert the equation. What if you forget the trigonometry formulas as well? Then I think you refer to my previous post about how to memorize trigonometry functions:http://tutapoint.com/knowledge-center/view/36/

Since sin2a+cos2a=1 , we get cos a = radical (1-sin2a) , and here a = arcsin x, so

(arcsinx)' = 1 / radical [1-sin2 (arcsin x)]

since  sin(arcsin x) = x, so sin(arcsin x) = x2    Therefore,

(arcsinx)' = 1 / radical (1-sinx2)

Now let's do (arctan x)' = ?

Similarly, we get

tan(arctan x) = x

Take derivative on both sides,

[tan(arctan x)]' = x'

[tan(arctan x)]' = 1

Use chain rule we get

sec2 (arctanx) * (arctan x)' = 1

(arctan x)' = 1 / [sec2(arctan x)]

Here again we use trigonometry formula  sec2a=1+ tan2a  , and  here a = arctan x , so

(arctan x)' = 1 / [1+ tan2(arctan x)]

(arctan x)' = 1 / (1+ x2)

Now let's do (arcsec x)' = ?

Again we get this :

sec (arcsec x) =  x

take derivatives on both sides we get

[sec (arcsec x)]' =  x'

[sec (arcsec x)]' = 1

By chain rule we get:

sec (arcsec x) * tan(arcsec x)   * (arcsec x)' = 1

(arcsec x)' = 1 / [sec (arcsec x) * tan(arcsec x)]

Since sec 2a = 1= tan 2a, tan a = radical (sec2a -1 ), and here a = arcsec x, so  tan(arcsec x) = radical [sec2 (arcsecx)  -1 ]  , and sec(arcsec x) =x. therefore now we get:

(arcsec x)' = 1 / [x * radical [sec(arcsecx)  -1]

Done.

(arcsinx)' = 1 / radical (1-sinx2)

(arctan x)' = 1 / (1+ x2)

(arcsec x)' = 1 / [x * radical [sec(arcsecx)  -1]

Interestingly,

we can add negative sign to all the 3 formulas above and get the other three formulas:

(arccosx)' = -1 / radical (1-sinx2)

(arccot x)' = -1 / (1+ x2)

(arccsc x)' = -1 / [x * radical [sec(arcsecx)  -1]

In conclusion, we can find 3 formulas by hand if we forget them, then add negative signs to get the remaining three formulas. That can help those people who often forget formulas and don't want to lose points.

• Subject : Math
• Topic : Calculus
• Posted By : Jason