Lösung 1.1:2f

Aus Online Mathematik Brückenkurs 2

We can rewrite the function using a trigonometric addition formula:

\displaystyle f\left( x \right)=\cos \left( x+\frac{\pi }{3} \right)=\cos x\centerdot \cos \frac{\pi }{3}-\sin x\centerdot \sin \frac{\pi }{3}

If we now differentiate this expression, \displaystyle \cos \frac{\pi }{3} and \displaystyle \sin \frac{\pi }{3} are constants and we obtain

\displaystyle \begin{align} & {f}'\left( x \right)=\frac{d}{dx}\left( \cos x\centerdot \cos \frac{\pi }{3}-\sin x\centerdot \sin \frac{\pi }{3} \right) \\ & =\cos \frac{\pi }{3}\centerdot \frac{d}{dx}\cos x-\sin \frac{\pi }{3}\centerdot \frac{d}{dx}\sin x \\ & =\cos \frac{\pi }{3}\centerdot \left( -\sin x \right)-\sin \frac{\pi }{3}\centerdot \cos x \\ \end{align}

If we then use the addition formula in reverse, this gives

\displaystyle \begin{align} & {f}'\left( x \right)=-\left( \sin x\centerdot \cos \frac{\pi }{3}+\cos x\centerdot \sin \frac{\pi }{3} \right) \\ & =-\sin \left( x+\frac{\pi }{3} \right) \\ \end{align}

NOTE: In the next section, we will go through differentiation rules which make it possible to differentiate the expression directly without rewriting in this way.