Aus Online Mathematik Brückenkurs 2

The entire expression is made up of several levels,

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } }

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

\displaystyle \cos \bbox[#FFEEAA;,1.5pt]{\phantom{\sqrt{\bbox[#FFCC33;,1.5pt]{1-x} } } }\,,

and differentiate this using the chain rule,

\displaystyle \frac{d}{dx}\,\cos \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} } = -\sin \bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\cdot \bigl(\bbox[#FFEEAA;,1.5pt]{\sqrt{1-x} }\,\bigr)'\,\textrm{.}

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

\displaystyle \bigl( \sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}\,\bigr)' = \frac{1}{2\sqrt{\bbox[#FFCC33;,1.5pt]{1-x}}}\cdot \bbox[#FFCC33;,1.5pt]{(1-x)}^{\,\prime}\,,

where we have used the differentiation rule,

\displaystyle \frac{d}{dx}\,\bigl(\sqrt{x}\,\bigr) = \frac{1}{2\sqrt{x}}\,\textrm{.}

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}\cdot\frac{d}{dx}\,\sqrt{1-x}\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot\frac{d}{dx}\,(1-x)\\[5pt] &= -\sin\sqrt{1-x}\cdot\frac{1}{2\sqrt{1-x}}\cdot (-1)\\[5pt] &= \frac{\sin\sqrt{1-x}}{2\sqrt{1-x}}\,\textrm{.} \end{align}