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!