Aus Online Mathematik Brückenkurs 2

The entire expression is made up of several levels,

\displaystyle \cos \left\{ \left. \sqrt{\left\{ \left. 1-x \right\} \right.} \right\} \right.

and when we differentiate we go from the outside inwards. In the first stage, we consider the expression as "cos of something",

\displaystyle \cos \left\{ \left. {} \right\} \right.

and differentiate this using the chain rule:

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

In the next differentiation, we have "the root of something",

\displaystyle \left( \sqrt{\left\{ \left. 1-x \right\} \right.} \right)^{\prime }=\frac{1}{2\sqrt{1-x}}\centerdot \left( 1-x \right)^{\prime }

where we have used the differentiation rule,

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

for the outer derivative.

The whole differentiation in one go becomes:

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