How to differentiate with respect to 'x' y=cos^(-1){2x √(1-x^(2))} ?

1 Answer
Jun 14, 2018

Long answer. Apologies if there is a shorter way; but this is the only way that I know to answer it to the best of my ability.

The answer I got was (2-4x^2)/(sqrt(1-5x^2 + 8x^4 -4x^6)

Explanation:

I believe the way to solve this is by using the chain rule. You will also need to know the standard differential of cos^-1 (x), and use the product rule on your substitution.

y = cos^-1(2x sqrt(1-x^2))

Step 1: Differential of 'u'
u = 2x sqrt(1-x^2) = 2x (1-x^2)^(1/2)

v = 2x (dv)/dx = 2
z = (1-x^2)^(1/2) (dz)/dx = -x(1-x^2)^(-1/2) (chain rule used here too)

(du)/dx = z*(dv)/dx + v*(dz)/dx
= 2(1-x^2)^(1/2) - 2x^2(1-x^2)^(-1/2)

Step 2: Differential of original function

y = cos^-1(u)

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

= (-1)/(sqrt(1-u^2)) * [2(1-x^2)^(1/2) - 2x^2(1-x^2)^(-1/2)]

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

And now to simplify the numerator:

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

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

=((2-4x^2)/sqrt(1-x^2))/(sqrt(1-(2x sqrt(1-x^2))^2))

And now to simplify the denominator:

=((2-4x^2)/sqrt(1-x^2))/(sqrt(1-(4x^2 (1-x^2))

=((2-4x^2)/sqrt(1-x^2))/(sqrt(1-4x^2 + 4x^4))

=(2-4x^2)/(sqrt(1-x^2)sqrt(1-4x^2 + 4x^4)

=(2-4x^2)/(sqrt((1-x^2)(1-4x^2 + 4x^4))

=(2-4x^2)/(sqrt(1-4x^2 + 4x^4 -x^2 +4x^4 -4x^6)

=(2-4x^2)/(sqrt(1-5x^2 + 8x^4 -4x^6)

Again, I hope that wasn't too confusing and would be keen to hear from others if there is a quicker and cleaner way of solving this.