Question

Find the derivatives of the following:

`tan^-1 = ((6x)/(1 - 9x^2))`

Answer 1

Let y = `tan^-1 ((6x)/(1 - 9x^2))`

y = `tan^-1 ((2 xx 3x)/(1 - (3x)^2))`  ........(1)

Put 3x = tan θ

(1) ⇒ y = `tan^-1 ((2 xx tan theta)/(1 - tan^2 theta))`

 y = `tan^-1 (tan 2theta)`

y = 2θ

`("d"y)/("d"theta) = 2  xx 1` = 2 ........(2)

3x = tan θ

`3 ("d"x)/("d"theta) = sec^2theta`

`3 ("d"x)/("d"theta) = 1 + tan^2 theta`

`("d"x)/("d"theta) = 1/3 (1 + (3x)^2)`

`("d"x)/("d"theta) = 1/3 (1 + 9x^2)`  .........(3)

Using equation (2) and (3) we have

`(("d"y)/("d"theta))/(("d"x)/("d"theta)) = 2/(1/3 (1 + 9x^2))`

`("d"y)/("d"x) = 6/(1 + 9x^2)`