Question

If y `tan^-1(sqrt((a - x)/(a +  x)))`, where – a < x < a, then `"dy"/"dx"` = .........

Options

• `x/sqrt(a^2 - x^2)`

• `a/sqrt(a^2 - x^2)`

• `-(1)/(2sqrt(a^2 - x^2)`

• `(1)/(2sqrt(a^2 - x^2)`

Answer 1

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

Explanation:

The derivative of tan⁡⁻¹(u) is given by:

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

Here u = `sqrt((a - x)/(a + x))`

Differentiate u `(a - x)/(a + x)

`u = sqrt((a - x)/(a + x))`

Let v = `(a - x)/(a + x)`, so u = `sqrtv`

The derivative of u is:

`(du)/(dx) = 1/(2sqrtv) . (dv)/(dx)`

Now differntiate v:

v = `(a - x)/(a + x)`

Using the quotient rule:

`(dv)/(dx) = ((a - x)(-1) - (a - x)(1))/(a + x)^2`

= `(- a - x - a + x)/(a + x)^2`

= `(-2a)/(a + x)^2`

Thus `(du)/dx = 1/(2sqrt((a - x)/(a + x))). (-2a)/(a + x)^2`

Substitute back into the formula

The derivative of y is:

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

Substitute `(du)/dx` and 1 + u² into the expression for `(dy)/dx`

`(dy)/dx = 1/((2a)/(a + x)) . (-a)/(sqrt((a - x)(a + x))/((a + x)))`

`(dy)/dx = (a + x)/2a . (-a)/((a + x)sqrt((a - x)(a + x)))`

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

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