# How do I find the derivative of y = arccos((x-3)^2)?

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 x + 6}{\sqrt{1 - {\left(x - 3\right)}^{4}}}$

#### Explanation:

The formula to find the derivative of ${\cos}^{-} 1 u$ is

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

So from the given $y = {\cos}^{-} 1 {\left(x - 3\right)}^{2}$
Let $u = {\left(x - 3\right)}^{2}$

from the formula

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

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

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

$- \frac{1}{\sqrt{1 - {\left(x - 3\right)}^{4}}} \cdot 2 \left(x - 3\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 3\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 \left(x - 3\right)}{\sqrt{1 - {\left(x - 3\right)}^{4}}} \cdot \left(1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 {\left(x - 3\right)}^{2}\right) = \frac{- 2 x + 6}{\sqrt{1 - {\left(x - 3\right)}^{4}}}$

have a nice day!