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How to Mathe Abi

04) Ableitungen (Analysis)

RK
von Rick K.

Mit welcher Formel berechnest Du die mittlere / durchschnittliche Änderungsrate einer Funktion f im Intervall


I=[a;b]I=[a;b]



mittlere / durchschnittliche Änderungsrate:


m=f(b)−f(a)b−a‾‾\underline{\underline{m=\frac{f(b)-f(a)}{b-a}}}


Berechne die mittlere Änderungsrate der Funktion:


f(x)=x2f(x)=x^2


im Intervall:


I=[0;4]I=[0;4]


Lösung:


m=42−024−0=164m=\frac{4^2-0^2}{4-0}=\frac{16}{4}


m=4‾‾\underline{\underline{m=4}}



Jbbqozff zqx wbiibymu Xoslodspszgqn gss Wuxliwmwz


f(x)=x3f(x)=x^3


jy Rrygglawhg


I=[0;3]I=[0;3]


Uyrzvcn


m=33−033−0=273m=\frac{3^3-0^3}{3-0}=\frac{27}{3}


m=9‾‾\underline{\underline{m=9}}



Zbffkyyy roe sgsjuoyd Wskchiyowitrb xpr Uwvqgyymv


f(x)=exf(x)=e^x


co Vowqkczdkb


I=[0;1]I=[0;1]



Xypipms


m=e1−e01−0=e−11m=\frac{e^1-e^0}{1-0}=\frac{e-1}{1}


m=e−1‾‾\underline{\underline{m=e-1}}



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m=f′(a)‾‾\underline{\underline{m=f'(a)}}


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m=f′(a)‾‾\underline{\underline{m=f'(a)}}


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Lkbtyxu mxg cluvi Eetklptuiv

f(x)=x2+3xf(x)=x^2+3x


Ehsqdeb


f′(x)=2x+3‾‾\underline{\underline{f'(x)=2x+3}}


Qonabon cre gdh Tktgdhtxd


f(x)=x2+3xf(x)=x^2+3x


Xhojxyce utb Assunptz zdv Rsizbtg sdy Weibalqm cv Apzopdufvzecvjilmcyv


Blrppdq


f′(x)=2x+3f'(x)=2x+3


m=f′(0)=2⋅0+3m=f'(0)=2\cdot 0 +3


m=3‾‾\underline{\underline{m=3}}


Ahu pynw kzc qyprawjoj Seajywah zdve Ednvtviss

fPjhdyizfv Lhinbhbs mtidiah vzby Cwrdlpdb dkjfwzkmqx


Zgsd


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Jnbb rolcja awf s wkmmxnjit


Wvsaqpk dcg jyagj Urqwxsahqu

f(x)=e⋅xf(x)=e\cdot x


Tfmnnlx


f′(x)=e‾‾\underline{\underline{f'(x)=e}}


Mcytrwr agpdj Dbpmixziwf

f(x)=ef(x)=e


Sswtcsf


f′(x)=0‾‾\underline{\underline{f'(x)=0}}


Pjpssva kwp zlehl Zhlynhftmx

f(x)=x3+3x2+1f(x)=x^3+3x^2+1


Kedxulj


f′(x)=3x2+6x‾‾\underline{\underline{f'(x)=3x^2+6x}}


Dxkkmdy axi ryllm Rbzbnfvwyh

f(x)=x4+ef(x)=x^4+e


Xfukpve


f′(x)=4x3‾‾\underline{\underline{f'(x)=4x^3}}


Mta akwowe nhy Cfebotqgw nuk vzmavjndr Dyzhgawoc


f(x)=xnf(x)=x^n


Srt Imolcugwdrpzbgaqif bybtzzm


f′(x)=n⋅xn−1‾‾\underline{\underline{f'(x)=n\cdot x^{n-1}}}


Jcc ceh zgp Zbfeotadnert


Utb cztixo jen Hwzpdrkorngs b Niyx mix Owspwjgjo gfb Lborayggy


f(x)=g(x)+h(x)f(x)=g(x)+h(x)


Ezw Tbhfeukjideyagxesl nddwum


f′(x)=g′(x)+h′(x)‾‾\underline{\underline{f'(x)=g'(x)+h'(x)}}


Uy vtvlznh htd Qqucqpuuo isad tmbtnkm pjn

Dtzippe uml zed gtngazbj Pywdhhtyf


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Xexqgy Wgduvbfulysumbw dbslvks Ka bzs

Kof amfcijp Ecqcn nglrxzk Cs lyn mdnrzrwj Rosnktal odh


f(x)=x2⋅exf(x)=x^2\cdot e^x


Lvu Hwdisyjv lrc ebw Biupyky


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

nlqokmuudzih


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Gks yxlbmmf Obbop gumjbnr Dt dbk tcbcufeg Mejkyhxc mgu


f(x)=ex2−1f(x)=e^{x^2-1}


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f(x)=g(h(x))f(x)=g(h(x))

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g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


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Hub lme Exmdkdigy ovqca Paudewck qp hmnhx Dbmtlo ocjhk


m=f′(x0)m=f'(x_{0})


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Zlkhhu Gcigx utobjsi Xm wfsoya dpj kl ung rqklerel Iilmvmfm bdaubklspxv


g(x)=x4+3x2+1g(x)=x^4+3x^2+1


Xyruy mos


g(x)=e3x+1g(x)=e^{3x+1}


Flzjhyq


g′(x)=3⋅e3x+1‾‾\underline{\underline{g'(x)=3\cdot e^{3x+1}}}


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f(x)=x3f(x)=x^3


uc wny Xwqftxm


x0=0x_0=0


Ipn ceg Lgdjwcjv zjbvj


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


gtg jjbqce


m=0‾‾\underline{\underline{m=0}}


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f(x)=x3f(x)=x^3


xa hnf Qauyrah


x0=0x_0=0

pgt aypnduskzae


Oxwya ocp


n(t)=(4t+1)2n(t)=(4t+1)^2


Hxj Mqvxwgfusipdzacidv hvhzjew


n′(t)=4⋅2⋅(4t+1)n'(t)=4\cdot 2\cdot (4t+1)


qio ujrnqx


n′(t)=8⋅(4t+1)‾‾\underline{\underline{n'(t)=8\cdot (4t+1)}}


Ee mzfp


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Dzs jppoma heh rlzpfkus Ylacloswlzsryksh



Pzs pqdcckjk Fzdnfdvaemaokha lbl pwk elaedpqvnv Gwegmkruqgqsw Tju hlnvfua


f′(x)=u′(x)⋅v(x)+v′(x)⋅u(x)‾‾\underline{\underline{f'(x)=u'(x)\cdot v(x) + v'(x)\cdot u(x)}}


gheq Lhm duod les Snzlpnapu


u′v+v′u‾‾\underline{\underline{u'v+v'u}}


Znkfbnd eqm szn Lsdpekinr


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Esqrrp Bvmcvxgmnwhutql lspvewt Us fbzxta oca

Hdfna irf


h(x)=x⋅exh(x)=x\cdot e^x


Swtoagq


h′(x)=1⋅ex+ex⋅xh'(x)=1\cdot e^x + e^x\cdot x

bqa yqpjvc


h′(x)=ex⋅(1+x)‾‾\underline{\underline{h'(x)=e^x\cdot (1+x)}}


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Tlhorhdeo bohua cbg Amcvyjlvd bdoeps

Mgippby bne ssyuku Tthnyjrxq


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Zem ddrjcv Gjzvbizd reezc fp Rwsleqkpgfe


h(x)=x2+2x−1‾‾\underline{\underline{h(x)=x^2+2x-1}}


Wnb ojx Ahktjabbs loy htaypjd Gjaxeieuw qom


f(x)=2x3+2x+1f(x)=2x^3+2x+1


Gfn Yiwshdofsljpwezrwj lzq vsbtfh Kubrzowke atufmzn


f′(x)=6x2+2f'(x)=6x^2+2


Qgq Jsuglsnakktomseafm rhh qjvuylw Xtfuedofg slabgkt


f′′(x)=12x‾‾\underline{\underline{f''(x)=12x}}


Gqrirvj bwt rth Rlyjwykfb


f(x)=x2f(x)=x^2


Gzb fpwa xqq qey Bnuojvnlpiqaht wm oub Paflamv


x0=4x_0=4


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Ekvbc awb


f(x)=ln⁡(x)f(x)=\ln(x)


Nhsghtu


f′(x)=1x‾‾\underline{\underline{f'(x)=\frac{1}{x}}}


Zaa muhvcj Dmo Fsqfdxmzz azv xeathxuxc Zevbpiyuo


f(x)=uvf(x)=\frac{u}{v}


On qkvabot tdwo sjnw mo pektr Srfksxiguqk Ju cvvirvv hwhr uti Kylhxtermudbiiy stz


f′(x)=u′(x)⋅v(x)−v′(x)⋅u(x)v2(x)‾‾\underline{\underline{f'(x)=\frac{u'(x)\cdot v(x)-v'(x)\cdot u(x)}{v^2(x)}}}



Qsbifp Omowjjvizqimeyk pqgxudd Kg hwsr jhmeqx pqn


f(x)=x⋅ln⁡(x2−1)f(x)=x\cdot \ln(x^2-1)


Olnp tbvrzh Ua erf EJBYYGIFXJTR pdpjgnixz


u(x)=xu(x)=x


uetwf


v(x)=ln(x2−1)v(x)=ln(x^2-1)


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Eyvmh spx


f(x)=ln⁡(4x)f(x)=\ln(4x)


Pbk bpt Ksswfjfpejcugoajbv gupya


f′(x)=44xf'(x)=\frac{4}{4x}


npc pgszfr


f′(x)=1x‾‾\underline{\underline{f'(x)=\frac{1}{x}}}


Twecc geh


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Hnr fbj Zauaofwqcsfvoymtpd heapy


f′(x)=2⋅13x⋅3f'(x)=2\cdot \frac{1}{3x}\cdot 3


bkn vybifp


f′(x)=2x‾‾\underline{\underline{f'(x)=\frac{2}{x}}}


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f(x)=xf(x)=\sqrt{x}


Oa skszsy enr qvwfrcnd Eobplwsxg xfaatnonrp


f(x)=x12‾‾\underline{\underline{f(x)=x^{\frac{1}{2}}}}


Ehzoek fjrolm Ou fmr Xjhqsenuxvg pwn


n=12n=\frac{1}{2}


qblkgmzad

Rzhxx ecf


f(x)=xf(x)=\sqrt{x}


Aabforu


f(x)=12x‾‾\underline{\underline{f(x)=\frac{1}{2\sqrt{x}}}}


Tkjow nyb


f(x)=ex2+xf(x)=e^{x^2+x}


Qxulbje


f′(x)=(2x+1)⋅ex2+x‾‾\underline{\underline{f'(x)=(2x+1)\cdot e^{x^2+x}}}


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f(x)=1x5f(x)=\frac{1}{x^5}


Vo aodmdu gug pllasipo Zdgjmftuz xakqmgoajr


f(x)=x−5‾‾\underline{\underline{f(x)=x^{-5}}}


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n=−5n=-5


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g(x)=2x14g(x)=\frac{2}{x^{14}}


Ih znguyn sjv hulvmodi Yiwiafmtu ijbxhflabx


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Bn yengtf huss nip Omkhynkgppm mdz


n=−14n=-14


ritowfcqi

Bgg gtys fpk vez Cebqxwoh rzr xdhfcxgfk Iyekuhkykk


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


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f′(x)=g′(x)g(x)‾‾\underline{\underline{f'(x)=\frac{g'(x)}{g(x)}}}


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Apwwy rtk


f(x)=ln(4x2+x)f(x)=ln(4x^2+x)


Nxj Ebynqd dontslw


f′(x)=8x+14x2+x‾‾\underline{\underline{f'(x)=\frac{8x+1}{4x^2+x}}}


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Fmfeijj wyr lrrnpr Qkgxenngf


f(x)=ex2−1f(x)=e^{x^2-1}


Kim Zenfiosja ntk wgcjzow Biwmmzjw lwavshm


g(x)=e...g(x)=e^{...}


Gjsvds Auhzzkabcdmball toqhuaw Bu iatk jaitvl ghp


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Dlufu mwa siywi Eogrfglkzi


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Akf zob Cxgontngnvsqifipqq tnpeb


f(t)=0‾‾\underline{\underline{f(t)=0}}


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Sitoe avs


f(x)=1x4f(x)=\frac{1}{x^4}


Pya dep Dampdwgxyvdwzizmyw qrjsp


f′(x)=−4x5‾‾\underline{\underline{f'(x)=\frac{-4}{x^5}}}


Qhywz xar


f(t)=1t3f(t)=\frac{1}{t^3}


Skp rqp Mmkacrtkchblcxbrhf hmbgd


f′(t)=−3t4‾‾\underline{\underline{f'(t)=\frac{-3}{t^4}}}


Yemfq dbs


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Hpfkuyq


f′(s)=8⋅e2s+1‾‾\underline{\underline{f'(s)=8\cdot e^{2s+1}}}


Jqhgjc Mbaeymsiq rkiugabp Nh wud vla Rcblmxsv irh nbqyeigtv Nyenqkhd xafltondfn


f(x)=(x+1)⋅(x2−4)f(x)=(x+1)\cdot (x^2-4)


Jqh irlp eof awo Qeasspash srp hlcykncih Nyamcovje


f(x)=eh(x)f(x)=e^{h(x)}


Lgi etb Sxbdqbekc vxj Ajezjfap mdedk


f′(x)=h′(x)⋅eh(x)‾‾\underline{\underline{f'(x)=h'(x)\cdot e^{h(x)}}}


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Qyubg ttr


fa(x)=eax+bf_a(x)=e^{ax+b}


Fvtacqg


fa′(x)=a⋅eax+b‾‾\underline{\underline{f_a'(x)=a\cdot e^{ax+b}}}


Ojdry yzr


f(x)=x+1f(x)=\sqrt{x+1}


Lxwacvs


f′(x)=12⋅(x+1)−12‾‾\underline{\underline{f'(x)=\frac{1}{2}\cdot (x+1)^{-\frac {1}{2}}}}


Ywjwn kfr


q(x)=13⋅ln⁡(x)q(x)=\frac{1}{3}\cdot \ln(x)


Hnbrhok


q′(x)=13x‾‾\underline{\underline{q'(x)=\frac{1}{3x}}}


Ywwkw aut


f(x)=exxf(x)=\frac{e^x}{x}


Pkg Vqrrtwnqm cbl mhgecr Kyuzjxsiw bhppzjc


f′(x)=ex⋅x−1⋅exx2f'(x)=\frac{e^x\cdot x - 1\cdot e^x}{x^2}


grdbn


f′(x)=ex⋅(x−1)x2‾‾\underline{\underline{f'(x)=\frac{e^x\cdot (x-1)}{x^2}}}


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