Den deriverte til funksjonar med fleire ledd

Vi deriverer ein funksjon multiplisert med ein konstant ved å derivere funksjonen og multiplisere med konstanten.

Døme 1

$$ $ \begin{array}{rcl}f\left(x\right)& = & 5x^2\\ & & \\ f^{\text{'}}\left(x\right)& =& 5·2x^{2-1}\\ & & \\ & =& 10x\end{array}$$$

Vi deriverer summar av og differansar mellom funksjonar ved å derivere ledd for ledd.

$$ \begin{array}{c}\begin{array}{c}\begin{array}{cc}& \end{array}\end{array}\end{array}f\left(x\right)=g\left(x\right)+h\left(x\right)\begin{array}{ccc}& \begin{array}{ccc}& \Rightarrow & \end{array}& \end{array}{f}^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)+h^{\text{'}}\left(x\right)$$

$$ $ \begin{array}{cc}\begin{array}{ccc}\begin{array}{ccc}& & \end{array}& & \end{array}& \end{array}f\left(x\right)=g\left(x\right)-h\left(x\right)\begin{array}{ccc}& \begin{array}{ccc}& \Rightarrow & \end{array}& \end{array}{f}^{\text{'}}\left(x\right)=g^{\text{'}}\left(x\right)-h^{\text{'}}\left(x\right)$$$

Døme 2

$$ \begin{array}{rcl}f\left(x\right)& = & 3+x^2\\ & & \\ f^{\text{'}}\left(x\right)& =& 0+2x^{2-1}\\ & & \\ & =& 2x\end{array}$$

Døme 3

$$ \begin{array}{rcl}g\left(x\right) & =& \frac{1}{3}{x}^3-2\sqrt{x}\\ & & \\ g\left(x\right) & =& \frac{1}{3}{x}^3-2·x^{\frac{1}{2}}\\ & & \\ g^{\text{'}}\left(x\right) & =& \frac{1}{3}·3·x^{3-1}-2·\frac{1}{2}·x^{\frac{1}{2}-1}\\ & & \\ & =& \frac{3}{3}·x^2-\frac{2}{2}·x^{-\frac{1}{2}}\\ & & \\ & = & x^2-\frac{1}{x^{{\displaystyle \frac{1}{2}}}}\\ & & \\ & =& x^2-\frac{1}{\sqrt{x}}\end{array}$$