Produktregelen for derivasjon

Første metode:

$\begin{array}{rcl}f\left(x\right) & =& x^3·x^5\\ & =& x^{3+5}\\ & =& x^8\\ f\text{'}\left(x\right) & =& 8x^{8-1}\\ & =& 8x^7\end{array}$

Andre metode:

$\begin{array}{rcl}f\left(x\right) & =& x^3·x^5\\ f\text{'}\left(x\right) & =& \left(x^3\right)\text{'}·x^5+x^3·\left(x^5\right)\text{'}\\ & =& 3x^2·x^5+x^3·5x^4\\ & =& 3x^{2+5}+5x^{4+3}\\ & =& 3x^7+5x^7\\ & =& 8x^7\\ & & \end{array}$

Første metode:

$\begin{array}{rcl}g\left(x\right) & =& \left(6x-1\right)\left(2x-2\right)\\ & =& 12x^2-12x-2x+2\\ & =& 12x^2-14x+2\\ g\text{'}\left(x\right) & =& 12·2x-14\\ & =& 24x-14\end{array}$

Andre metode:

$\begin{array}{rcl}g\left(x\right) & =& \left(6x-1\right)\left(2x-2\right)\\ g\text{'}\left(x\right) & =& {\left(6x-1\right)}^{\text{'}}·\left(2x-2\right)+\left(6x-1\right)·{\left(2x-2\right)}^{\text{'}}\\ & =& 6·\left(2x-2\right)+\left(6x-1\right)·2\\ & =& 12x-12+12x-2\\ & =& 24x-14\end{array}$

Husk at her skal vi derivere med hensyn på $z$. Første metode:

$\begin{array}{rcl}h\left(z\right) & =& {\left(x-5z\right)}^2\\ & =& x^2-2x5z+25z^2\\ h\text{'}\left(z\right) & =& -10x+50z\end{array}$

Andre metode:

$\begin{array}{rcl}h\left(z\right) & =& \left(x-5z\right)·\left(x-5z\right)\\ h\text{'}\left(z\right) & =& \left(x-5z\right)\text{'}·\left(x-5z\right)+\left(x-5z\right)·\left(x-5z\right)\text{'}\\ & =& -5·\left(x-5z\right)+\left(x-5z\right)·\left(-5\right)\\ & =& -5x+25z-5x+25z\\ & =& -10x+50z\end{array}$

$\begin{array}{rcl}f\left(x\right) & =& (5x^3+4)(x-1)\\ f\text{'}\left(x\right) & =& (5x^3+4)\text{'}·(x-1)+(5x^3+4)(x-1)\text{'}\\ & =& 15x^2(x-1)+(5x^3+4)·1\\ & =& 15x^3-15x^2+5x^3+4\\ & =& 20x^3-15x^2+4\end{array}$

$\begin{array}{rcl}g\left(x\right) & =& (3x^2+2)(\sqrt{x}-1)\\ g\text{'}\left(x\right) & =& (3x^2+2)\text{'}(\sqrt{x}-1)+(3x^2+2)(\sqrt{x}-1)\text{'}\\ & =& 6x(\sqrt{x}-1)+(3x^2+2)·\frac{1}{2\sqrt{x}}\\ & =& \frac{6x(\sqrt{x}-1)h·2\sqrt{x}+(3x^2+2)}{2\sqrt{x}}\\ & =& \frac{12x^2-12x\sqrt{x}+3x^2+2}{2\sqrt{x}}\\ & =& \frac{15x^2-12x\sqrt{x}+2}{2\sqrt{x}}\end{array}$

$\begin{array}{rcl}h\left(x\right) & =& (x^2+3)·x^3\\ h\text{'}\left(x\right) & =& (x^2+3)\text{'}·x^3+(x^2+3)·(x^3)\text{'}\\ & =& 2x·x^3+(x^2+3)·3x^2\\ & =& 2x^4+3x^4+9x^2\\ & =& 5x^4+9x^2\end{array}$

I denne oppgaven kan det bli uoversiktlig hvis vi fører slik som vi har gjort i de tidligere oppgavene. Her kan det være hensiktsmessig å se på de to faktorene hver for seg først.

$\begin{array}{rcl}j\left(x\right)& = & \left(\frac{1}{3}{x}^3-\frac{2}{x^2}\right)·3x^{-2}\\ u\left(x\right)& =& \frac{1}{3}{x}^3-\frac{2}{x^2}=\frac{1}{3}{x}^3-2x^{-2}\\ u\text{'}\left(x\right)& =& 3·\frac{1}{3}{x}^{3-1}-2·\left(-2\right)·x^{-2-1}=x^2+4x^{-3}\\ v\left(x\right)& =& 3x^{-2}\\ v\text{'}\left(x\right)& =& 3·\left(-2\right)·x^{-2-1}=-6x^{-3}\\ j\text{'}\left(x\right)& =& u\text{'}\left(x\right)·v\left(x\right)+u\left(x\right)·v\text{'}\left(x\right)\\ & =& \left(x^2+4x^{-3}\right)·3x^{-2}+\left(\frac{1}{3}{x}^3-2x^{-2}\right)·\left(-6x^{-3}\right)\\ & =& 3x^{2-2}+12x^{-3-2}-2x^{3-3}+12x^{-2-3}\\ & =& 1+24x^{-5}\\ & & \end{array}$

$\begin{array}{rcl}f\left(x\right) & =& {\left(2x-2\right)}^2\\ f\left(2\right) & =& {\left(2·2-2\right)}^2=(4-2{)}^2\\ & =& 2^2=4\end{array}$

Vi finner den deriverte til funksjonsuttrykket i punktet der $x=2$. Dermed finner vi stigningen til tangenten som berører grafen til f i punktet $\left(2,f\left(2\right)\right)$:

$\begin{array}{rcl}f\text{'}\left(x\right) & =& (2x-2)\text{'}·(2x-2)+(2x-2)·(2x-2)\text{'}\\ & =& 2·(2x-2)\text{'}·(2x-2)\\ & =& 2·2·(2x-2)=8x-8\\ f\text{'}\left(2\right) & =& 8·2-8=8\end{array}$

Vi finner likningen til tangenten:

$\begin{array}{rcl}y-y_1 & =& a\left(x-x_1\right)\\ y-4 & =& 8(x-2)\\ y & =& 8x-16+4\\ y & =& 8x-12\end{array}$

$\begin{array}{rcl}g\left(x\right) & =& 2x\left(x^2-x\right)\\ & =& 2x^3-2x^2\\ g\left(2\right) & =& 2·2^3-2·2^2\\ & =& 16-8\\ & =& 8\end{array}$

Vi finner den deriverte til funksjonsuttrykket i punktet der $x=2$. Dermed finner vi stigningen til tangent som berører grafen til f i punktet $(\mathrm{2,} g(2\left)\right)$:

$\begin{array}{rcl}g\text{'}\left(x\right) & =& 2·\left(x^2-x\right)+2x\left(2x-1\right)\\ & =& 2x^2-2x+4x^2-2x\\ & =& 6x^2-4x\\ g\text{'}\left(2\right) & =& 6·2^2-4·2=16\end{array}$

Vi finner likningen til tangenten:

$\begin{array}{rcl}y-y_1 & =& a\left(x-x_1\right)\\ y-8 & =& 16(x-2)\\ y & =& 16x-32+8\\ y & =& 16x-24\end{array}$