# How do you find the derivative of  y= sqrt(x^2 + cos x) using the chain rule?

Dec 14, 2015

y'= (2x-sinx)/(2sqrt(x^2+cosx)

#### Explanation:

Given $\text{ } y = \sqrt{{x}^{2} + \cos x}$
Rewrite : $\text{ } y = {\left({x}^{2} + \cos x\right)}^{\frac{1}{2}}$

Using both power and chain rule to differentiate this function

$y ' = \frac{1}{2} {\left({x}^{2} + \cos x\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + \cos x\right)$

$\implies \frac{1}{2} {\left({x}^{2} + \cos x\right)}^{- \frac{1}{2}} \cdot \left(2 x - \sin x\right)$

=>(2x-sinx)/(2sqrt(x^2+cosx)