Question

Differentiate the following w.r.t. x: 

`sin^-1(sqrt((1 + x^2)/2))`

Answer 1

Let y = `sin^-1(sqrt((1 + x^2)/2))`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[sin^-1(sqrt((1 + x^2)/2))]`

= `(1)/(sqrt(1 - (sqrt((1 + x^2)/2)))^2)."d"/"dx"(sqrt((1 + x^2)/2))`

= `(1)/(sqrt((1 - (1 + x^2)/2))^2) . 1/(2sqrt((1 + x^2)/2)) . 1/2 . 2x`

= `(1)/(((sqrt(2 - 1 + x^2))/sqrt2)^2) . 1/((2sqrt(1 + x^2))/sqrt2) . 1/2 . 2x`

= `(1)/((sqrt(1 - x^2)/sqrt2)^2) . 1/((2sqrt(1 + x^2))/sqrt2) . 1/2 . 2x`

= `(1)/(((1 - x^2)/2)) . 1/(2((sqrt(1 + x^2))/sqrt2)) . 1/2 . 2x`

= `2/((1 - x^2)) . sqrt2/(2sqrt(1 + x^2)) . 1/2 . 2x`

= `(2 . sqrt2 . 2x)/((1 - x^2) . 2 . 2 . sqrt(1 + x^2))`

= `(sqrt2 . x)/((1 - x)^2 (sqrt(1 + x^2))`

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

= `x/sqrt(1 - x^4)`

Answer 2

Substitution

Let x² = cos(2θ). This is a useful identity because `(1+cos(2theta))/2 = cos^2(theta)`

Now substitute this into your equation:

`y = sin^-1 (sqrt((1+cos(2theta))/2))`

Since `(1+cos(2theta))/2 = cos^2(theta)`

`y = sin^-1 (sqrt(cos^2(theta)))`

y = sin^–1 (cos(θ))

Use Trigonometric Identities

`costheta = sin (pi/2-theta)`

`y = sin^-1(sin(pi/2 - theta))`

`y = pi/2 - theta`

Back-substitute θ

x² = cos(2θ)

2θ = cos^–1 (x²)

`theta = 1/2 cos^-1 (x^2)`

`y = pi/2 - 1/2 cos^-1 (x^2)`

Differentiate

Now differentiate with respect to x:

The derivative of `pi/2` is 0

The derivative pf `cos^-1(u) is - 1/sqrt(1-u^2) * (du)/dx`

`dy/dx = 0 - 1/2 (- 1/sqrt(1-(x^2)^2) * d/dx (x^2))`

`dy/dx = 1/2 * 1/sqrt(1-x^4) * (2x)`

The 2s cancel out, leaving you with:

`dy/dx = x/sqrt(1-x^4)`