Produktregelen for derivasjon

Døme 1

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

Her kunne vi òg ha multiplisert ut parentesane før vi deriverte.

Døme 2

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

Kan vi multiplisere ut parentesen her òg før vi deriverer?

Bevis for produktregelen

$\begin{array}{ccl}f\text{'}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{f(x+∆x)-f\left(x\right)}{∆x}\\ & =& \underset{∆x\to 0}{\lim }\frac{u\left(x+∆x\right)·v\left(x+∆x\right)-u\left(x\right)·v\left(x\right)}{∆x}\end{array}$

Vi legg så til og trekker frå det same uttrykket i teljaren.

$\begin{array}{ccl}f\text{'}\left(x\right)& =& \underset{∆x\to 0}{\lim }\frac{u\left(x+∆x\right)·v\left(x+∆x\right)\stackrel{=0}{-\overbrace{u\left(x\right)·v(x+∆x)+u\left(x\right)·v(x+∆x)}}-u\left(x\right)·v\left(x\right)}{∆x}\\ & & \\ & =& \underset{∆x\to 0}{\lim }\frac{(u\left(x+∆x\right)-u(x\left)\right)·v(x+∆x)}{∆x}+\underset{∆x\to 0}{\lim }\frac{u\left(x\right)·\left(v\right(x+∆x)-v\left(x\right))}{∆x}\\ & & \\ & =& \stackrel{u\text{'}\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }\frac{(u\left(x+∆x\right)-u(x\left)\right)}{∆x}}}·\stackrel{v\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }v\left(x+∆x\right)}}+\stackrel{u\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }u\left(x\right)}}·\stackrel{v\text{'}\left(x\right)}{\overbrace{\underset{∆x\to 0}{\lim }\frac{\left(v\right(x+∆x)-v\left(x\right))}{∆x}}}\\ & & \\ & =& u\text{'}\left(x\right)·v\left(x\right)+u\left(x\right)·v\text{'}\left(x\right)\end{array}$

Film: Produktregelen for derivasjon

I filmen under (lengde 6:53) får du ein gjennomgang av døme 1 over.