# How do you differentiate sqrt(cos(x^2+2))+sqrt(cos^2x+2)?

Apr 18, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x s e n \left({x}^{2} + 2\right) + s e n \left(x + 2\right)}{\sqrt{\cos} \left({x}^{2} + 2\right) + \sqrt{{\cos}^{2} \left(x + 2\right)}}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\cos} \left({x}^{2} + 2\right) + \sqrt{{\cos}^{2} \left(x + 2\right)}} \cdot s e n \left({x}^{2} + 2\right) \cdot 2 x + 2 s e n \left(x + 2\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x s e n \left({x}^{2} + 2\right) + 2 s e n \left(x + 2\right)}{2 \sqrt{\cos} \left({x}^{2} + 2\right) + \sqrt{{\cos}^{2} \left(x + 2\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cancel{2} \left(x s e n \left({x}^{2} + 2\right) + s e n \left(x + 2\right)\right)}{\cancel{2} \sqrt{\cos} \left({x}^{2} + 2\right) + \sqrt{{\cos}^{2} \left(x + 2\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x s e n \left({x}^{2} + 2\right) + s e n \left(x + 2\right)}{\sqrt{\cos} \left({x}^{2} + 2\right) + \sqrt{{\cos}^{2} \left(x + 2\right)}}$