Aus Online Mathematik Brückenkurs 2

When we see this expression, we should think "root of something",

\displaystyle \sqrt{\left\{ \left. {} \right\} \right.}

and in order to differentiate it, we should first differentiate the outer function , "the root of", with respect to its argument and, after that, multiply by the derivative of the inner functional expression

\displaystyle \left\{ \left. {} \right\} \right.=\cos x,

\displaystyle \frac{d}{dx}\sqrt{\left\{ \left. \cos x \right\} \right.}=\frac{1}{2\sqrt{\left\{ \left. \cos x \right\} \right.}}\centerdot \left( \left\{ \left. \cos x \right\} \right. \right)^{\prime },

where we have used the differentiation rule

\displaystyle \frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}.

Thus, we obtain

\displaystyle \frac{d}{dx}\sqrt{\cos x}=\frac{1}{2\sqrt{\cos x}}\centerdot \left( -\sin x \right)=-\frac{\sin x}{2\sqrt{\cos x}}