# How do you find the derivative of the function: sqrt(1-9x^2)arccos(3x)?

Jul 21, 2017

$y = \sqrt{1 - 9 {x}^{2}} \arccos \left(3 x\right)$

Let $\text{ } 3 x = \cos \theta$

So Differentiating w r to x we have $\text{ } 3 = - \sin \theta \frac{d \theta}{\mathrm{dx}}$

$\implies \frac{d \theta}{\mathrm{dx}} = - \frac{3}{\sin} \theta$

Now

$y = \sqrt{1 - 9 {x}^{2}} \arccos \left(3 x\right)$

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

$\implies y = \theta \sin \theta$

Differentiating w r to $\theta$ we have

$\frac{\mathrm{dy}}{d \theta} = \theta \cos \theta + \sin \theta$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} \times \frac{d \theta}{\mathrm{dx}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\theta \cos \theta + \sin \theta\right) \times \left(- \frac{3}{\sin} \theta\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \left(- 3 \theta \cos \frac{\theta}{\sin} \theta - 3\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 \left(\frac{\theta \cos \theta}{\sqrt{1 - {\cos}^{2} \theta}} + 1\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 \left(\frac{3 x {\cos}^{-} 1 \left(3 x\right)}{\sqrt{1 - 9 {x}^{2}}} + 1\right)$