Question

If y = `tan^-1((6x - 7)/(6 + 7x))`, then `dy/dx` = ______.

Options

• `6/(1 + (6 + 7x)^2`

• `7/(1 + (6 + 7x)^2`

• `1/(1 + x^2)`

• `6/(1 + x^2)`

Answer 1

If y = `tan^-1((6x - 7)/(6 + 7x))`, then `dy/dx` = `underlinebb(1/(1 + x^2))`.

Explanation:

y = `tan^-1((6x - 7)/(6 + 7x))`

By dividing numerator and denominator by 6 we get

∴ y = `tan^-1((x - 7/6)/(1 + (7x)/6))`

∴ y = `tan^-1x - tan^-1  7/6`   ...`[∵ tan^-1((a - b)/(1 + ab)) = tan^-1a - tan^-1b]`

∴ `dy/dx = 1/(1 + x^2) - 0 = 1/(1 + x^2)`