5. Derivatives of inverse trigonometric functions

Derivatives of inverse trigonometric functions:

Inverse of sin x = arcsin(x) or

Let us now find the derivative of Inverse trigonometric function

Example: Find the derivative of a function 

Solution: Given

Differentiating the above equation w.r.t. x, we have:

 

Putting the value of y form (i), we get

From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined  

 

i. e.  

From (i) we have

 

Using property of trigonometric function,

Now putting the value of (iii) in (ii), we have

Therefore, the Derivative of Inverse sine function is

Example:Find the derivative of a function

Problem: y = cot^-1(1/x²)

Solution:

As we are solving the above three problem in the same way this problem will solve

By using chain rule,

y’ = (cot^-1(1 / x²))’

= { – 1 / (1 + (1 / x²))² } . (1 / x²)’

= { – 1 / (1 + (1 / x⁴)) . (-2x^-3)

= 2x⁴ / (x⁴ + 1)x³

Example: Solve f(x) = tan^-1(x) Using first Principle.

Solution: 

For solving and finding tan^-1x, we have to remember some formulae, listed below.

• lim_h->0 {f(x + h) – f(x)} / h
• tan^-1(θ/θ) = 1
• tan^-1x – tan^-1y = tan^-1[(x – y) / (1 + xy)]

f(x) = tan^-1x

f(x + h) = tan^-1(x + h)

Apply 1st formula

lim_h->0 {tan^-1(x + h) – tan^-1x } / h

Now Apply 3rd formula

lim_h->0 tan^-1[(x – h – x) / (1 + (x + h)x] / h

lim_h->0 tan^-1[(h  / (1 + x² + xh ] / h . [(1 + x² + xh) / (1 + x² + xh)]

lim_h->0 tan^-1 {h / 1 + x² + xh} / {h / 1 + x² + xh} . lim_h->0 1 / 1 + x² + xh

Now we made the solution like so that we apply the 2nd formula

= 1 . 1 / (1 + x² + x . 0)

= 1 / (1 + x²)