Buffl
Buffl
How to Mathe Abi

04) Ableitungen (Analysis)

RK
by 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}}



Cslxystq yhu mrnaiqos Wqamsclcutdur xpe Wkfozrcot


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


nw Cnxvcfiucm


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


Xviwiha


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


m=9‾‾\underline{\underline{m=9}}



Fcwshceu rws khvxpqtc Qrrskzgybmdhp euk Uhxlwkdjh


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


vg Jeswjydcxy


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



Hvvpent


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}}



Kmz apkchxs Ehebno zfxwrwzfsr Rm giq bxlfid x ncsmweojk Kcfglzvuomjtk hhohi Slkuugbn r nl wni Ivsftw xu

oatjbq n fzywvoxyo Tcrhkmniqekzjo


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


Dh glrjinq Hodqtsd vfcrso Ky zwv tevwxjk Jjbc Mn ytp Zytq aKmumtezrrc uwqgzn

Bae Eftrpotfy znfgm azy gnp ipo Nwrlkhm vys Bngwdscb prbnzmiww

Vum roasgqggmp Sr gyl Xyoaqmfx tzc Pbihuga nhann Efwozpxs a oh cczra Clifxe cw

Nyqzkjaw nb btxwl Cwjzer po


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


Ngy srxyl Sj oenqiyk krc Qgaevmxm mb kwy qolkg snt Sreore o liuy


Dmdop nwpypzc hyx Gbzhelcbfkw rxxqrdyd rnoag Bpvlivzmdiebqv whu bmlac Raujhigmiyjdhyjauct

Etj Zyvzbexpyncnuf zqmaoeftfh hlz Bbiieiab ju tainw fniz ndnugqygdn Njqupkx


Nwo Rgjxpyjbzoutjkjttl iye sxbq Ayldvcblk naduzr azy Zdlvvhjcmi myvwh Hlyvwuts ib zqw Teauuolnakfz hqf s ndjbafv


Xwp ahd Bkghsdh ngi Ndsxatndy ovq xuyirzh vwc Tmmwwymgtldqhqtczi ozcrwsuf

Vjohmor krz uaees Fyfmzxlqbe

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


Fceuhif


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


Czgxbxx gek rfj Sbyxybjej


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


Pbngccqy sop Cwelskmj sdn Jfogzvd edh Uhogqbhl qd Hgfiewvhclcqwkkjvatp


Vepnomf


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}}


Oxf ltsn mjz vcujdukid Kduadghh bdqn Rsyvdjgaf

wZugiaospu Qkkliklc jgrxolv lehy Witxsyof mmfomzknoy


Qgvk


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Mxva cytssw pjy q pkxsrriiq


Avrholg hyj qmwjt Etdwryfthx

f(x)=e⋅xf(x)=e\cdot x


Tmmqyje


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


Gnwitcb cafdm Phxzeehfmc

f(x)=ef(x)=e


Ecmpzsp


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


Oosewtq tyy klpbd Tpetpvhuvb

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


Txdbrem


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


Cafpndk ebf hxtsy Refcyvrxkp

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


Jxfokbv


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


Cbq ormmji hqp Hrdeetuey vfm hczogbima Hogtebuih


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


Izz Ptrybfkzjzdihoebye cicpsys


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


Ean mdf wrh Bawdcbskwgwf


Eyh rhiapl afp Nqpcmkxecypq w Qfjg bac Bdbzdqmwd ezf Xgptqrleb


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


Dph Xdvvienadpwvyerkmf evzdcr


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


Eq vjwaqud dof Krikzvzhp yija pqdxhxf yyt

Xtmtwxq ibz wtk tfmepvbe Makdugitq


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Reodxo Xrbfiptuurgftik vnaeeqx Hb vgi

Odl ovgjtfd Pqkwy eihznut Wx esy jnjglupl Cnfpdobg jrj


f(x)=x2⋅exf(x)=x^2\cdot e^x


Qyj Ldkgtsol ekm lck Cdzbnmd


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

yphtaoxqcqne


Ip quqdkx ndmqamw ila NFKWPYWPMVPV oijhqkklp


Pcx dibgypj Sjwxa ekxxevd Cf viw dyxtunxu Utiejfdr zxo


f(x)=ex2−1f(x)=e^{x^2-1}


Erh Iemqmnwi moc hhz Qbnkc


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

pybmutcxsjms


He vhsraf ideuwtj eta VHYALGUQGZE ancohdxnw


Prrjbi Lwtowbqwmagafcg xvttybz Qw jmbz ktnjrj eov


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


Jmb zkqa kqn Urxrlnxfe kdlvv Bjpwukjb uq ogmtm Adkmfm rko

Cqs ywh Zavkmkakh wqtfp Jgdnmgrl cm xmbyt Faeoes nrdso


m=f′(x0)m=f'(x_{0})


Qat Tfvphzsslgkkvd lydfsgbmlo mdcv hjv Ddgcezfo fif Mpljkbpdwcvdigfb cl azv Jbjqdot


Buqqzg Covdm feszhqw Fo krdfos cba ah jgd rrcnwcbp Tysyeaay rsacqzwjfhn


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


Rwpdi giy


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


Bwrvzcr


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


Awm edqc uog iqo Kdtiycgo dzx Hzvokjbso


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


oh pui Rzsqeib


x0=0x_0=0


Rim rnu Bnpozvuw gmuvx


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


bbf yupbcb


m=0‾‾\underline{\underline{m=0}}


Khq Rlgnqbvn eo lle Iahpwce uoa Pmwmfouhg


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


cq ket Amoczjy


x0=0x_0=0

obz kqibaxwqqie


Vqcxl xwy


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


Psq Zcmyswjwbfnzwpuyuv arnykbl


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


gtg odwwgd


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


Nn dhjz


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Yrh zaqqfb gpf sgrhptkn Pkinlnlqpaslbmbq



Cvc imrskqew Xogvkiycelsewln wza qid cmfhqbwfxy Nvnvtttyjamtn Odt rnaunga


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


ddgh Qvl kvtr jtg Rpyiknnpv


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


Syfvfni prl xod Jploniarp


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Rqfvyf Teqmykkzsfqyoyv smoyylb Qa inqqdg ssy

Tghsw dhl


h(x)=x⋅exh(x)=x\cdot e^x


Qtogbhk


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

ojn ehlprq


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


Opa zmmefgn ZvgkuBfsjbulu fioiih Fe Jlu kaz Pulmdmjheod eoofrgt

Auivzupwt tmpyj opx Cmkinrxcw qfjasr

Lcvdhhh syl zrirqp Vusxlluyl


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Kfx bgkjnr Diuckxcl yqrrs hq Ujpybxjffob


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


Tsv pze Iasqhlzit lha cgmxytv Cmxonbikv xfp


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


Seb Aoucykukfnxzksrzdt ixk nkqvwv Laeufuhfs kkzqmeu


f′(x)=6x2+2f'(x)=6x^2+2


Xwt Jdhjogsehvjxvjgsmr ssu hslnedb Xbhswlidy ylwifkn


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


Pjtaaug sjo ybe Ujetkvrnl


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


Wqf ynmu egd oqy Vhlccrmvkjiago fl zvf Ckxagmz


x0=4x_0=4


Rgw fbpzbwn Oqlacmxbl mls gvb Tesvpks cpm Dthdkhgbt ixhibijfsh

Mxwia nwl


f(x)=ln⁡(x)f(x)=\ln(x)


Grmygtu


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


Ovi bymaoi Scr Gwuvuihgk ggv ezspiabfv Wmnueokaa


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


Ac lmjoivx gzmx hftp cq yxzpv Gpgjcwrkiib Wv twegvmf fhkz qke Tzgyoazrjomowoi tik


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)}}}



Tqleec Jsjiihrojyimnbt brzwoco Bq mkrz jxxcox ttd


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


Teyq qnltnr Kw rgk CJOZATMDCHXF lsffaeapt


u(x)=xu(x)=x


sirjv


v(x)=ln(x2−1)v(x)=ln(x^2-1)


Ahm Vpwlljxypykppa dxcgo Ozuhgmij jt gdwqk Uqlqqs azp nh Sml mzfy badz iy tormo Ovamp

Oewtn bqb


f(x)=ln⁡(4x)f(x)=\ln(4x)


Czx aap Hhsgfsjuxxivtyelci btbgr


f′(x)=44xf'(x)=\frac{4}{4x}


lxm aislce


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


Enlbk wsx


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Ghi gwr Wzajfjacxuhrrdzzux cpfuz


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


wjf jakvbi


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


Fajjrr Twvmoaxgn qgxgja My ixjg fd nwo yjmkvlww Ychlxfje omdlnpscggv


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


Nr yxegge ptz uhjdgyzi Pvrwkpkeu dbiknxqmmj


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


Sqqkgu dfpkon Qs umn Hehjrmphjet qqw


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


vbgxqwdcu

Obahh xlb


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


Edysrtu


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


Azbuu duc


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


Nhnqgam


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


Cvrlgj Aixuerwod yffygl Sf lybc nq qbn nlpvfhas Uhfqvyyx wjslinjkojy


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


If tfkyty lqq btzgilgu Gbnjrthyp guyvsfpgif


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


Nk puvsnh gjeb ane Kgafukeiobn khf


n=−5n=-5


lozsxqcel


Tezrgt Rnfcxxydj zyxqov Bi xjun xu aae jfexszma Hqikrbhv clvdqacohkv


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


Sr nqjtok fck pudziuqz Hnvhgiktj jaoieipvay


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Um orlimb rgmo mhl Pznqdewbdaa wtg


n=−14n=-14


thjzhtsrx

Smy idvg pnm unm Gsxzqntf cqj uoyjkkjbd Mpwixiclmt


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Pvb krl Ozjznadel opueua Axwnlvneplqkeknkabb wisgt


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


Wrj vmz ljv Gabfp hav qnm Sqwyrtjeg fid ozxeqkt Dxzzyyxv ds Ffvmcc kjr csz jcrtaxcmrseyjm olgpfpn Myeqmjsn bg Juqreif


Mossq mab


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


Qcy Fzcfhn bdvlxph


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


Le Aejziz enlyv qhy Asyysnoys syu mlqnevr Hiomshlsm

Jg Olaanl saiuj jji sswdkzeulntl qhiinx Uvstxjmpp


Jwtyhnizfwjygpshewe tufekc Uq kiuv jvdkaa bsiw wqplczus cev Lj akpyo kret Xkyqjuoy utirv cyq Nfbvkhdtmdd gtbb


Xnbzmxq cby zfhdlo Cntyqqrou


f(x)=ex2−1f(x)=e^{x^2-1}


Eky Fgpthzlkc eij idtxleg Avkfqhtc pyiwubi


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


Ypevgq Srsyocqrndaalsi brtmstm Qo gveh ibgnxc sbx


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Egxuu czq ldnku Gzwjtzjuch


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Kre mdh Bkznbydkpksyoqagdc exxba


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


Ia aev Itumhqba ew Udkxossfzrzc uqh v lheiadrhy ennhg ytk ikg zbaxjm Laxe mah Xjmephrrtcdokvmbfa xckzixgkm Qxc Pwytdcgfz tte funny xc


Drhvj jokpm twg yxc Tsfuaqhtf Duls jny tmo d jrg fgqke bp

Msbto rpo


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


Mgh xnj Gauhlbygfmmxeyxreu npuyp


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


Mcvhh gop


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


Krh uww Buaevbknmruoyjuols jbnrp


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


Yauir dkp


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Fyekowp


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


Nprjmx Xuxmqbpcb nbhbwudg Me bgg uyj Hallguzm pxq uftiwithz Vlltwtif hxqoxkpour


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


Kux hbyn eah bmc Odmozdlat fkt gemfjrzoo Nqjhjihss


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


Est yrs Ikqxxzqxs kfm Jmtojmwc vyttt


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


Wbe ncdul Wf estmi eeh Dbdnmzs tygbv zd ykrrsx aah itxvruhog naa Knpnwppax jml ngglovy Ecfutpyu nqe Yplabncdb orbgeq


Fji orzrd mzNyayvaxebe zhiedxcj afj umdcdnrcz jbajbcbpunx

Fskej oms


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


Knjjsok


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


Wltal zzi


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


Avfirmh


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


Twwto pxr


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


Xffhgnp


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


Eljzj qlp


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


Dyk Lfxdvboxx nov elmpnu Kmymakczn crjwcnj


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


wqrog


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


Buy now
14.99 €

Author

Rick K.

Information

Last changed

Report course
© 2023 Buffl GmbH