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

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 ${\cos}^{-} 1 \left(x\right)$, and use the product rule on your substitution.

$y = {\cos}^{-} 1 \left(2 x \sqrt{1 - {x}^{2}}\right)$

Step 1: Differential of 'u'
$u = 2 x \sqrt{1 - {x}^{2}} = 2 x {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

$v = 2 x$ $\frac{\mathrm{dv}}{\mathrm{dx}} = 2$
$z = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$ $\frac{\mathrm{dz}}{\mathrm{dx}} = - x {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$ (chain rule used here too)

$\frac{\mathrm{du}}{\mathrm{dx}} = z \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{dz}}{\mathrm{dx}}$
$= 2 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} - 2 {x}^{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$

Step 2: Differential of original function

$y = {\cos}^{-} 1 \left(u\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\sqrt{1 - {u}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(2 x \sqrt{1 - {x}^{2}}\right)$

$= \frac{- 1}{\sqrt{1 - {u}^{2}}} \cdot \left[2 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} - 2 {x}^{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}\right]$

$= \frac{2 {\left(1 - {x}^{2}\right)}^{\frac{1}{2}} - 2 {x}^{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}}{\sqrt{1 - {\left(2 x \sqrt{1 - {x}^{2}}\right)}^{2}}}$

And now to simplify the numerator:

$= \frac{2 \sqrt{1 - {x}^{2}} - \frac{2 {x}^{2}}{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {\left(2 x \sqrt{1 - {x}^{2}}\right)}^{2}}}$

$= \frac{\frac{2 \left(1 - {x}^{2}\right) - 2 {x}^{2}}{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {\left(2 x \sqrt{1 - {x}^{2}}\right)}^{2}}}$

$= \frac{\frac{2 - 4 {x}^{2}}{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - {\left(2 x \sqrt{1 - {x}^{2}}\right)}^{2}}}$

And now to simplify the denominator:

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

$= \frac{\frac{2 - 4 {x}^{2}}{\sqrt{1 - {x}^{2}}}}{\sqrt{1 - 4 {x}^{2} + 4 {x}^{4}}}$

=(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.