3. Derivatives of the Inverse Trigonometric Functions

by M. Bourne

Recall from when we first met inverse trigonometric functions:

"sin-1x" means "find the angle whose sine equals x".

Example 1

If x = sin-10.2588 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-1x.

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-1x 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.

Continues below

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-15x.

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 − x2).

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(x2 + 3y).

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 + (x2 + 3y)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(x2 + 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)`

2. Find the derivative of y = 4 tan-13x4.

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)`

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3. Find the derivative of y = (x2 + 1) sin-14x.

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`

4. Find the derivative of sin-1(x + y) + y = x2.

Answer

We have an implicit function:

sin-1(x + y) + y = x2.

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`