Show `d/dx (cos^-1 2x)=-2/(1-4x^2)^(1/2).` If `y=sin (cos^-1 2x),`show that `(1-4x^2) (d^2y)/(dx^2)-4x dy/dx+4y=0.` How to do this?

We want to show that:

`d/dx cos^(-1)(2x)=-2/(1-4x^2)^(1/2)`
`y =sin( cos^(-1)(2x)) => (1-4x^2) (d^2y)/(dx^2)-4x dy/dx+4y=0`

We start with the standard calculus result:

`d/dx cos^(-1)x = -1/sqrt(1-x^2)`

Then applying the chain rule, we have:

`d/dx cos^(-1)(2x) = -1/sqrt(1-(2x)^2) d/dx (2x)`
`" " = -2/sqrt(1-4x^2) \ \ \ ` QED

so then, given `y =sin( cos^(-1)(2x))`, then applying the chain rule, we have:

`dy/dx = cos( cos^(-1)(2x)) d/dx cos^(-1)(2x)`
`\ \ \ \ \ = (2x) (-2/sqrt(1-4x^2))`
`\ \ \ \ \ = (-4x) / sqrt(1-4x^2)`

And then by applying the quotient rule:

`(d^2y)/(dx^2) = { (sqrt(1-4x^2))(d/dx -4x) - (-4x)(d/dx sqrt(1-4x^2)) } / (sqrt(1-4x^2))^2`

`\ \ \ \ \ \ = { (sqrt(1-4x^2))(-4) + 4x(1/2(1-4x^2)^(-1/2)d/dx(-4x^2)) } / (1-4x^2)`

`\ \ \ \ \ \ = { (sqrt(1-4x^2))(-4) + 4x(1/2(1-4x^2)^(-1/2)(-8x)) } / (1-4x^2)`

`\ \ \ \ \ \ = { -4sqrt(1-4x^2) - (16x^2)/sqrt(1-4x^2) } / (1-4x^2)`

And so the LHS of the given DE is:

`LHS = (1-4x^2) (d^2y)/(dx^2)-4x dy/dx+4y`

`\ \ \ \ \ \ \ \ = (1-4x^2) {{ -4sqrt(1-4x^2) - (16x^2)/sqrt(1-4x^2) } / (1-4x^2)} -4x {(-4x) / sqrt(1-4x^2)} +4sin( cos^(-1)(2x))`

`\ \ \ \ \ \ \ \ = -4sqrt(1-4x^2) - (16x^2)/sqrt(1-4x^2) +(16x^2) / sqrt(1-4x^2) +4sin( cos^(-1)(2x))`

`\ \ \ \ \ \ \ \ = 4sin( cos^(-1)(2x)) -4sqrt(1-4x^2)`

Now if we use the trigonometric identity `sin^2 A + cos^2A -=1`, then:

`(sin (cos^(-1)2x))^2 + (cos(cos^(-1)2x))^2 -=1`

`:. (sin (cos^(-1)2x))^2 =1 - (cos(cos^(-1)2x))^2`
`:. sin (cos^(-1)2x) =sqrt(1 - (2x)^2) = sqrt(1 - 4x^2)`

So, we have:

`LHS = 4{sqrt(1 - 4x^2)} -4sqrt(1-4x^2)`
`\ \ \ \ \ \ \ \ = 0 \ \ \ ` QED

We differentiate using the Chain Rule :

`d/dx(cos^-1 2x)`,

`=-1/sqrt{1-(2x)^2}*d/dx(2x)`,

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

`rArr d/dx(cos^-1 2x)=-2/(1-4x^2)^(1/2)`, as desired!

For ease of writing, we will use the following notations :

`dy/dx=y_1 and (d^2y)/dx^2=d/dx(dy/dx)=d/dx(y_1)=y_2`.

Note that : `(1) : d/dx(y^2)=2y*dy/dx=2yy_1`.

`(2) : d/dx(y_1^2)=2y_1*d/dx(y_1)=2y_1y_2`.

Next, we have, `y=sin(cos^-1 2x).........(star_1)`.

`:. dy/dx=cos(cos^-1 2x)*d/dx(cos^-1 2x)`.

`:. y_1=cos(cos^-1 2x)*{-2/sqrt(1-4x^2)}..."[from above]"`.

`:. -1/2sqrt(1-4x^2)y_1=cos(cos^-1 2x)......(star_2)`.

Hence, squaring and adding `(star_1) and (star_2)`, we get,

`y^2+1/4(1-4x^2)y_1^2=1, or,`

`4y^2+(1-4x^2)y_1^2=4`.

Rediff.ing w.r.t. `x` and using `(1) & (2)` above, we get,

`4(2yy_1)+{(1-4x^2)(2y_1y_2)+y_1^2(-8x)}=0`.

Dividing throughout by `2y_1!=0` & rearranging, we get,

`(1-4x^2)y_2-4xy_1+4y=0`, as desired!

Enjoy Maths!