Produktregelen for derivasjon

Vi har en egen regel for å finne den deriverte av et produkt av to funksjoner:

$$ \begin{array}{cccccl}f\left(x\right)=u\left(x\right)·v\left(x\right)& & \begin{array}{ccc}& \Rightarrow & \end{array}& & & f^{\text{'}}\left(x\right)=u^{\text{'}}\left(x\right)·v\left(x\right)+u\left(x\right)·v^{\text{'}}\left(x\right)\\ & & & & & \\ f=u·v& & \begin{array}{ccc}& \Rightarrow & \end{array}& & & f^{\text{'}}=u^{\text{'}}·v+u·v^{\text{'}}\end{array}$$

$f$, $u$ og $v$ er funksjoner av $x$ og skal deriveres med hensyn på $x$. I den andre linja ovenfor har vi brukt en litt forenklet skrivemåte.

$$ \begin{array}{rcl}f\left(x\right)& = & \stackrel{u\left(x\right)}{\overbrace{\left(x^2+3x\right)}}·\stackrel{v\left(x\right)}{\overbrace{\left(x^3+1\right)}}\\ & & \\ f^{\text{'}}\left(x\right)& =u^{\text{'}}& \left(x\right)·v\left(x\right)+u\left(x\right)·v^{\text{'}}\left(x\right)\\ & & \\ f^{\text{'}}\left(x\right)& =& \stackrel{u\text{'}\left(x\right)}{\overbrace{\left(2x+3\right)}}·\stackrel{v\left(x\right)}{\overbrace{\left(x^3+1\right)}}+\stackrel{u\left(x\right)}{\overbrace{\left(x^2+3x\right)}}·\stackrel{v\text{'}\left(x\right)}{\overbrace{3x^2}}\\ & & \\ f^{\text{'}}\left(x\right)& =& 2x^4+2x+3x^3+3+3x^4+9x^3\\ & & \\ f^{\text{'}}\left(x\right)& =& 5x^4+12x^3+2x+3\end{array}$$

Her kunne vi også ha multiplisert ut parentesene før vi deriverte.

$$ \begin{array}{rcl}f\left(x\right)& =& \stackrel{u\left(x\right)}{\overbrace{(x-1)}}·\stackrel{v\left(x\right)}{\overbrace{\sqrt{x}}}\\ & & \\ f^{\text{'}}\left(x\right)& =& \stackrel{u\text{'}\left(x\right)}{\overbrace{1}}·\stackrel{v\left(x\right)}{\overbrace{\sqrt{x}}}+\stackrel{u\left(x\right)}{\overbrace{\left(x-1\right)}}·\stackrel{v\text{'}\left(x\right)}{\overbrace{\frac{1}{2\sqrt{x}}}}\\ & & \\ f^{\text{'}}\left(x\right)& =& \frac{\sqrt{x}·2\sqrt{x}}{2\sqrt{x}}+\frac{\left(x-1\right)}{2\sqrt{x}}\\ & & \\ f^{\text{'}}\left(x\right)& =& \frac{2x+x-1}{2\sqrt{x}}\\ & & \\ f^{\text{'}}\left(x\right)& =& \frac{3x-1}{2\sqrt{x}}\end{array}$$

Kan vi multiplisere ut parentesen her også før vi deriverer?

$$\begin{gathered} \begin{array}{ccl}{f}^{\text{'}}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{u\left(x+∆x\right)·v\left(x+∆x\right)-u\left(x\right)·v\left(x\right)}{∆x}\\ & & \\ & & \text{Vi legger så til og trekker fra det samme uttrykket i telleren.}\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{u\left(x+∆x\right)·v\left(x+∆x\right)\stackrel{=0}{\overbrace{-u\left(x\right)·v(x+∆x)+u\left(x\right)·v(x+∆x)}}-u\left(x\right)·v\left(x\right)}{∆x}\\ & & \\ f^{\text{'}}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{(u\left(x+∆x\right)-u(x\left)\right)·v(x+∆x)}{∆x}+\underset{∆x\to 0}{\lim }\frac{u\left(x\right)·\left(v\right(x+∆x)-v\left(x\right))}{∆x}\\ & & \\ f^{\text{'}}\left(x\right)& =& \stackrel{u\text{'}\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }\frac{(u\left(x+∆x\right)-u(x\left)\right)}{∆x}}}·\stackrel{v\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }v\left(x+∆x\right)}}+\stackrel{u\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }u\left(x\right)}}·\stackrel{v^{\text{'}}\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }\frac{\left(v\right(x+∆x)-v\left(x\right))}{∆x}}}\\ & & \\ f^{\text{'}}\left(x\right)& =& u^{\text{'}}\left(x\right)·v\left(x\right)+u\left(x\right)·v^{\text{'}}\left(x\right)\\ & & \end{array} \\ \end{gathered}$$