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Basic rules of derivatives
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "RULES
\nRule | \nFunction, $f(x)$ | \nDerivative, $f'(x)$ | \n
Power | \n$x^r$ | \n$r \\cdot x^{r-1}$ | \n
Sum/Difference | \n$u(x)+v(x)$ | \n$u'(x)+v'(x)$ | \n
Constant Multiplier | \n$k \\cdot f(x)$ | \n$k \\cdot f'(x)$ | \n
Product | \n$u \\cdot v$ | \n$u' \\cdot v + u \\cdot v'$ | \n
Quotient | \n$\\displaystyle {\\frac u v}$ | \n$\\displaystyle {\\frac {u' \\cdot v - u \\cdot v'} {v^2}}$ | \n
Chain | \n$\\displaystyle{g(u(x))}$ | \n$\\displaystyle{g'(u(x)) \\cdot u'(x)}$ | \n
1) Den deriverte av $\\;\\simplify {f(x) = {a} x + {b}}\\;$ finner vi ved å kombinere reglene (1), (2) og (3), men vi kan også se at grafen til funksjonen er en rett linje med konstant stigningstall {a}, og det blir verdien av den deriverte.
Etter regel (1) og (3) blir den deriverte av en konstant $b$ lik null fordi vi kan skrive $b = b \\cdot 1 = b \\cdot x^0$ , og deriverer vi $f(x)=x^0$ etter regel (1) får vi $f'(x) = 0 \\cdot x^{-1} = 0$ - det skulle bare mangle, en konstant har en linje parallell med x-aksen som graf, med stigningstall null. Den deriverte av $\\; {f(x) = \\var{a} x = \\var{a} x^1}\\;$ blir etter produktregel$\\;{f'(x) =\\var{a} x = \\var{a} \\cdot 1 \\cdot x^0 = \\var{a}}\\;$
Kort sagt, den deriverte av $\\;f(x)=a \\cdot x + b\\;$ er $\\;f'(x)=a\\;$.
2) Funksjonsuttrykket $\\;\\simplify {f(x) = {a1} x^2 + {b1}x + {c1}}\\;$ er en sum av tre ledd som deriveres hver for seg, $\\;f'(x)= \\var{a1} \\cdot (x^2)' + \\var{b1} \\cdot (x)' + \\var {c1} \\cdot (x^0)' = 2 \\cdot \\var{a1} x^1 + 1 \\cdot \\var {b1} \\cdot x^0 + 0 \\cdot \\var{c1} \\cdot x^{-1} = \\var {a1} x + \\var{b1}$
\n3) Den deriverte av $\\;{f(x) = \\frac {x^\\var{a2}} {\\var{b2}} - \\frac {\\var{b2}} {x^{\\var{a2}}} = \\frac 1 {\\var{b2}} \\cdot x^\\var{a2} - \\var {b2} \\cdot x^{- \\var{a2}} }\\;$ deriveres som $\\;f'(x) = \\frac 1 {\\var{b2}} \\cdot \\var {a2} \\cdot x^{\\var{a2} - 1} - \\var {b2} \\cdot (- \\var{a2}) \\cdot x^{-\\var {a2}-1}$
\n5) Her kan produktregelen brukes, men hvorfor ikke utvide uttrykket først?
$f(x) = (\\var{a}-x^{\\var{b}})x^\\frac {1} {\\var{n}} = \\var{a}x^\\frac {1} {\\var{n}} - x^{\\simplify{{b}+1/{n}}}$
6) Brøkfunksjonen $\\;f(x) = \\frac {\\simplify{x^{n}-{a1}}}{\\simplify{{n} - x^{n2}}} = \\frac u v\\;$ har $\\;u' = \\var {n} \\cdot x^{\\var{n}-1}\\;$ og $\\;v' = - \\var {n2} \\cdot x^{\\var{n2}-1}\\;$ som settes inn i brøkregelen, $\\;(\\frac u v)' = \\frac {u' \\cdot v - u \\cdot v'} {v^2}$
\n7) Funksjonen $\\;f(x) = \\var{a2}(x^{\\var{b2}}-\\var{n})^\\var{n2} =\\var{a2}(u)^\\var{n2} \\;$ har $\\; u = x^{\\var{b2}}-\\var{n}\\;$ som kjernefunksjon, med derivert $\\;u' = \\var {b2} \\cdot x^{\\var{b2}-1}\\;$
Den deriverte blir $\\;f'(x) = \\var{a2} \\cdot \\var {n2} \\cdot (x^{\\var{b2}}-\\var{n})^{\\var{n2}-1} \\cdot \\var {b2}\\cdot x^{\\var{b2}-1}\\;$ .
1)The derivative of $\\;\\simplify {f(x) = {a} x + {b}}\\;$ is $\\;f'(x)\\;$ = [[0]]
\n2) The derivative of $\\;\\simplify {f(x) = {a1} x^2 + {b1}x + {c1}}\\;$ is $\\;f'(x)\\;$ = [[1]]
\n3) The derivative of $\\;{f(x) = \\frac {x^\\var{a2}} {\\var{b2}} - \\frac {\\var{b2}} {x^{\\var{a2}}}}\\;$ is $\\;f'(x)\\;$ = [[2]]
\n4) The derivative of $\\;f(x) = x^\\frac {\\var{t}} {\\var{n}}\\;$ is $\\;f'(x)\\;$ = [[3]]
\n5) The derivative of $\\;f(x) = (\\var{a}-x^{\\var{b}})x^\\frac {1} {\\var{n}}\\;$ is $\\;f'(x)\\;$ = [[4]]
\n6) The derivative of $\\;f(x) = \\frac {\\simplify{x^{n}-{a1}}}{\\simplify{{n} - x^{n2}}}\\;$ is $\\;f'(x)\\;$ = [[5]]
\n7) The derivative of $\\;f(x) = \\var{a2}(x^{\\var{b2}}-\\var{n})^\\var{n2}\\;$ is $\\;f'(x)\\;$ = [[6]]
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