# 3. Derivatives of the Inverse Trigonometric Functions

by M. Bourne

Recall from when we first met inverse trigonometric functions:

"sin^{-1}x" means "find the angle whose sine equalsx".

### Example 1

If *x* =
sin^{-1}0.2588 then by using the calculator, *x* =
15°. We have found the angle whose sine is 0.2588.

### Notation

We also write: arcsin *x* to mean the same
thing as sin^{-1}*x*.

It is better to use arcsin *x* because normally in mathematics, a number raised to the power `-1` means the reciprocal. Example: `3^-1=1/3`.

Most calculators use the (confusing) notation: `sin^-1 x`. This section mostly uses the sin^{-1}*x* notation (since it was originally written to be consistent with calculator notation), however you are encouraged to use the superior notation, `arcsin x`.

You may also wish to go back to background information on inverse trigonmetric equations.

## Derivatives of Inverse Trigonometric Functions

The following are the formulas for the derivatives of the inverse trigonometric functions:

`(d(sin^-1u))/(dx)=1/sqrt(1-u^2)(du)/(dx)`

`(d(cos^-1u))/(dx)=(-1)/sqrt(1-u^2)(du)/(dx)`

`(d(tan^-1u))/(dx)=1/(1+u^2)(du)/(dx)`

### Example 2

Find the derivative
of *y* = cos^{-1}5*x*.

Answer

Put `u = 5x` so `y = cos^-1 u`.

`(dy)/(dx)=(-1)/(sqrt(1-u^2))(du)/(dx)`

`=(-1)/(sqrt(1-(5x)^2))(d(5x))/(dx)`

`=(-5)/(sqrt(1-25x^2))`

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### Example 3

Find the derivative
of *y* = sin^{-1}(1 − *x*^{2}).

Answer

Put `u = 1 - x^2`

Then we have: `y = sin^-1 u`

`y=sin^-1(1-x^2)`

`(dy)/(dx)=1/sqrt(1-u^2)(du)/(dx)`

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

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

Expanding the expression under the square root and simplifying gives:

`(dy)/(dx)=(-2x)/(sqrt(-x^4+2x^2)`

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### Example 4

Find `(dy)/(dx)` if *x *+ *y* = tan^{-1}(*x*^{2} + 3*y*).

Answer

This is an **implicit function.**

`x+y=tan^-1(x^2+3y)`

`1+(dy)/(dx)` `=1/(1+(x^2+3y)^2)(2x+3(dy)/(dx))`

Multiplying throughout by 1 + (*x*^{2} + 3*y*)^{2} gives:

`(1+(x^2+3y)^2)+` `(1+(x^2+3y)^2)(dy)/(dx)` `=2x+3(dy)/(dx)`

`(dy)/(dx)(1+(x^2+3y)^2 -3)` `=2x-(1+(x^2+3y)^2)`

`(dy)/(dx)=` `(2x-1-(x^2+3y)^2)/(-2+(x^2+3y)^2)`

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## Exercises

1. Find the derivative of *y* = 3 cos^{-1}(*x*^{2} + 0.5).

Answer

Put `u = x^2+ 0.5`.

Then we have: `y = 3\ cos^-1 u`

`y=3\ cos^-1(x^2+0.5)`

`(dy)/(dx) =3[(-1)/(sqrt(1-u^2))(du)/(dx)]`

`=3[(-1)/(sqrt(1-(x^2+0.5)^2))(d(x^2+0.5))/(dx)]`

`=3[(-1)/(sqrt(1-(x^2+0.5)^2))(2x)]`

`=(-6x)/(sqrt(1-(x^2+0.5)^2)`

Expanding and simplifying gives:

`(dy)/(dx)=(-6x)/(sqrt(0.75-x^4-x^2)`

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2. Find the derivative of *y* = 4 tan^{-1}3*x*^{4}.

Answer

Put `u = 3x^4` and we have: `y = 4\ tan^-1 u`.

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=4(1/(1+u^2))(du)/(dx)`

`=4 1/(1+(3x^4)^2)(12x^3)`

`=(48x^3)/(1+9x^8)`

3. Find the derivative of *y* = (*x*^{2} + 1) sin^{-1}4*x*.

Answer

`y=(x^2+1)sin^-1 4x`

Let `u=x^2+1` and `v=sin^(-1) 4x`.

Then

`(dy)/(dx)=u (dv)/(dx)+v (du)/(dx)`

`=(x^2+1)[1/(sqrt(1-(4x)^2)) (4)]+` `(sin^(-1) 4x)(2x)`

`=(4(x^2+1))/(sqrt(1-16x^2))+2x\ sin^(-1)4x`

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4. Find the derivative of sin^{-1}(*x *+ *y*) + *y *= *x*^{2}.

Answer

We have an **implicit** function:

sin

^{-1}(x+y) +y=x^{2}.

Taking the first term, sin^{-1}(*x *+ *y*), and letting

u=x+y,

we differentiate the inverse sine using:

`1/sqrt(1-mu^2)(du)/(dx)`

and `(du)/dx = 1 + dy/dx`.

Using the above, and differentiating implicitly term-by-term gives:

`1/sqrt(1-(x+y)^2)(1+(dy)/(dx))+(dy)/(dx)` `=2x`

Multiplying throughout by:

`sqrt(1-(x+y)^2)`

we have:

`(1+(dy)/(dx))+sqrt(1-(x+y)^2)(dy)/(dx)` `=2xsqrt(1-(x+y)^2`

Subtracting 1 from both sides:

`(dy)/(dx)+sqrt(1-(x+y)^2)(dy)/(dx)` `=2xsqrt(1-(x+y)^2)-1`

Grouping the `dy/dx` terms:

`(1+sqrt(1-(x+y)^2))(dy)/(dx)` `=2xsqrt(1-(x+y)^2)-1`

Dividing both sides by:

`1+sqrt(1-(x+y)^2)`

We obtain the required solution:

`(dy)/(dx)=(2xsqrt(1-(x+y)^2)-1)/(1+sqrt(1-(x+y)^2`

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