$\begin{array}{rcl}f(x)& =& x^2\cdot \cos (x)\end{array}$

Wende die Produktregel an.

$\begin{array}{rcl}{f}^{\prime }(x)& =& 2x\cdot \cos (x)+x^2\cdot (-\sin (x))\\ & =& 2x\cdot \cos (x)-x^2\cdot \sin (x)\end{array}$

Schreibe den Bruch in ein Produkt um.

$\begin{array}{rcl}f& =& {\displaystyle \frac{\sin x}{x}}\\ & & \\ & =& \sin x\cdot {\displaystyle \frac{1}{x}}\\ & & \\ & =& \sin x\cdot x^{-1}\end{array}$

Wende die Produktregel an.

$\begin{array}{rcl}{f}^{\prime }(x)& =& \cos x\cdot x^{-1}+\sin x\cdot (-1)\cdot x^{-2}\\ & & \\ & =& {\displaystyle \frac{\cos x}{x}}-{\displaystyle \frac{\sin x}{x^2}}\\ & & \\ & =& {\displaystyle \frac{1}{x}}\cdot (\cos x-{\displaystyle \frac{\sin x}{x}})\\ & & \\ & =& {\displaystyle \frac{x\cdot \cos x-\sin x}{x^2}}\end{array}$

$f\left(x\right)=\sin \left(x\right)\cdot \cos \left(x\right)$

Wende die Produktregel an.

$\begin{array}{rcl}{f}^{\prime }(x)& =& \cos x\cdot \cos x+\sin x\cdot (-\sin x)\\ & =& {\cos }^{2}x-{\sin }^{2}x\end{array}$