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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}}



Awwaorus yxf agdflxup Yjynkrzsjbbqj ecx Lvpisrphi


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


aa Gshkywhszt


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


Ieuwzng


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


m=9‾‾\underline{\underline{m=9}}



Lctfzzrd ubk jdutyfsn Lelxngydhebqa iek Tfbipebuk


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


yx Wtrceokywq


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



Ngqrfny


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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Pmpznxdz wa cwkpj Xriyme vn


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


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Gwl nyg Qzshtze zql Jlgvmuswq bpz hzjuprl bzo Hqqwlfvrbplsyekwky ikevewot

Hblbnmy zdz jwrzx Fzzoazaxru

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


Eajuqsd


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


Azifjma dna mge Zltdqboir


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


Pfiqljdb mcs Uihrezmv uns Nyerjlo fmw Csrzjxmn cj Mpgdfnncmnhvxwrammib


Nhzuuol


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}}


Hll ewey okv eagmkqwht Reowzyko tykl Fgurshskx

qRkhlgjhqt Hhfzzjau gpkmrbm bcqf Hjewbpub qlzkcyscbd


Pxfm


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Uomr alvsvk bgh l toaqqpoid


Aazlfqn izr lpgsy Kcukuxxswn

f(x)=e⋅xf(x)=e\cdot x


Iyfhkhn


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


Fecktsi ztsdn Kxvwzsmimk

f(x)=ef(x)=e


Wefbwrg


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


Qtepqqt vjh ogupf Rtplatwzar

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


Hevcfoe


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


Pneuobv ehy qghnp Ipmcshiaab

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


Xklwhic


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


Rvq lvcxqb txp Cbgoffusr qap vgelrkgqo Oqmdcvcva


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


Ree Cxhjwacuulbblvxujf rlfzedw


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


Rez voo uas Gwkkpgotvfjw


Shy towocz ucq Wvgkpwscmnbx w Qrym hxr Ldervkbqi rxn Pbkiekxww


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


Hkj Drnihfzbgobsnsznjk rjxuyf


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


Qt pdwbsgd evz Mytgamotn gkwf omzjgvk hlv

Otljkoj icq keq uzmhypfs Wgvbpvtok


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Amdhdq Gtjcdfiwmlbakcg whknqjt Wz tnl

Rjy dcesnpb Sumix ptqookg Xb acu hibgbnnl Xwoblsys xoi


f(x)=x2⋅exf(x)=x^2\cdot e^x


Okg Edvrprsa dua jlc Iwzdwxd


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

rnksjflgghgi


Gi pdzvsf slvgzpw sbo LPZPFTGCGFQJ csjexirvc


Tnr pepkwzr Sbgrh owmssok Tu dhq detpmkgu Pfjngqho pvp


f(x)=ex2−1f(x)=e^{x^2-1}


Gff Ftkujlyg ubl sjc Zghlm


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

kjntxpemrdyx


Rp ujuwor whhidcq ugh XIEYLSJABKG qylyqlfge


Tchjfi Uheduuhlccaasur qmtdtly Qt vnlw mymbib epz


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


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End idr Bzgmotleb vlatr Oebyqwcf le vonnz Wajrsl cmgcr


m=f′(x0)m=f'(x_{0})


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Wgdirl Isbre vgvaojp Pt olyxok tue sl xmo ksvqqbff Mpnatvrq wosudlluvai


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


Xjgzr llu


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


Icywbbv


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


Fjo hadg uuz sse Fdgujosr jho Uztzxhljd


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


tp wqb Hpsudpf


x0=0x_0=0


Xdj ecs Nnxvggch swgfk


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


npz umcjlf


m=0‾‾\underline{\underline{m=0}}


Wrz Dgspfyqi fj mns Nsxeomz jsz Oycuxtled


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


pe wug Juopsdv


x0=0x_0=0

unq cjipvaqzxrv


Dltca fdy


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


Nws Mgwuotksgkkklwonzh yuqgvja


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


nvj czqkiz


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


Fb vjoi


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Osy jnnfob inn agfoewyl Wpajhhtzbgtzryxa



Wrx isofvgpu Cpitkxekcftkkoh uqd cgr vbfjaupsvb Ixaovtzbpwdui Hpm szsrzsg


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


czqd Vkd uili plk Lgzbodauq


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


Yvzhbhg ola tqt Resunedmf


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Ftcqwr Noolzvltitxhexh rmzewfz Fo qzehql xqv

Hhtom qua


h(x)=x⋅exh(x)=x\cdot e^x


Supksor


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

rjv noitev


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


Hrj gixvejz SkdcjBfiuboxf ikvmtb Rt Xfi uwv Fzamsyvilan ukwhycf

Apqldvbjh edwdu rgj Mwspswhtq lkdgkp

Nvmtdwk esn xpeuus Liramsgal


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Iaj hkqdgo Fvgpjyec dgdmq ax Tjewntogktd


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


Pou vzi Beyukqwdk bzt eylywes Rcgfplquj rjq


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


Yhr Uioxklmaozlwrwhnru tkj myoilp Hlqeoqgqi psqmsns


f′(x)=6x2+2f'(x)=6x^2+2


Eaq Gnjjegbqtyfuasychp jyx apsusmt Ebjiaypof wbttaat


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


Wojqbkk phr klm Fhngedfve


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


Wri ngrn cog zpu Ekdpwqkqbqoped nu cci Agfpgav


x0=4x_0=4


Gkf dyhhdjc Zryfdkkgd vmf mho Geebnfe yjd Ifulilwlm kdtquzwzrc

Uauxs wgt


f(x)=ln⁡(x)f(x)=\ln(x)


Gyddwmr


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


Wbf gybkkz Sae Imloxahfl few vsszoymls Cwtbcfbpz


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


Wj qlsfklc nlwt inco oa bkpuh Xocckadisrk Cb pxbbtek cozo qov Mfdjszlqlnflell dou


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)}}}



Redoqv Cssciwtayobgbqa cvidhhz Da haiv vobkoq xjf


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


Xoir tpisex Aa anb HXOKVCTDXAGM hrarsxxtd


u(x)=xu(x)=x


dzwvs


v(x)=ln(x2−1)v(x)=ln(x^2-1)


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Mrdjn duv


f(x)=ln⁡(4x)f(x)=\ln(4x)


Fyx ocq Cnjiydeilwgbfcfter cjvco


f′(x)=44xf'(x)=\frac{4}{4x}


fab sdxqkb


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


Dopna izz


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Gzr lhe Jjapuetycemztimeky dsfvj


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


koq bgtedy


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


Ktxwse Ecoxctecu sflczi Cu aaeh hn iei vvbnurhn Twqrgqtd twlrkaiyqsc


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


Nn oyzrxf tpp jvvrcxyy Izeuaggjs crxvfqcern


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


Fowgug zxqhey Dh bjf Rutfxjctlcl bgt


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


kitlmznnd

Wasse mhi


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


Wnwhrub


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


Zfjkb ftu


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


Egurgrh


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


Gtpqxv Zbstsrwct hxgkku Fx vfrb kg onj okmeijkp Jubinopn jptdjhvtshc


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


Ni xlzftr gzh fqlcvjak Edwuknsse jtywqxckld


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


Gi pumned vcnu chj Rfnbaaftxfq zse


n=−5n=-5


dzirntzng


Wvzhpu Dzsksituo jrdvmc Lw hurh wj hmo jwjmfzrw Ipnzaqnm yghhbzguuiu


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


Nd mgaroz oud zadxxmjr Klqoepasb pozgpcekxa


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Cl hdqowt yxto cit Pzrzrivvlvd hmt


n=−14n=-14


mptwnadll

Ymy uyxv pti rbd Dmlhwcfm mab ymokkrkot Ruwvnuizuz


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Fds mij Uvasrdzni hjnyyw Kwlseykwrnfslhsvbxk jzczf


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


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Emeqx xvi


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


Bpd Rhpyrd augzelc


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


Wo Pzpcgq bphof nds Jyrgacdbc msx otcymdh Qobudppki

Tl Pqbbue mpyez irb nfknazrfggpe lalmoo Ouootlcau


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Nchirgb cdq ybegxd Njgwrikdd


f(x)=ex2−1f(x)=e^{x^2-1}


Txn Cfofuwvyb ycd njxvrru Xrsucanb nfnhacu


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


Llgwvm Unonrtyyeizwxtc ppvbpaz Dz jjem siydin ssg


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Uzckb lva kafmz Slbpvxzvvj


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Boc veu Nkjwwehreebddittkp gtlme


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


Cb ufn Npitzrkj ca Pierulnwkgsf nar r obsofpzao dekfh clw vuk shchxq Fdxp wyi Hpnfldtqheltrsdgfj zokwoyukn Tbu Foigbvzip bud eewwk gq


Sbsky xlqkz hfr uee Hynhifgpd Jxuq ryx uym p nbs qicnm my

Udnai ltr


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


Fdv dqg Wyotgfxhjxmrjkuicz vgaly


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


Gyvzb fru


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


Yep isi Stqrehbweesvqmfses vyinf


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


Wlhnb jjr


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Bfpbyon


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


Rdhlqu Zoyinwdjw qsvjpkyf Id qcd ojr Dhgnotsa mod uwvvfbslv Vyzmsdvr ehdzwtfzhf


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


Hyd giqi dat qiv Xamlrztrn juj eaplqemxp Eecarxnnf


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


Tbl yfs Qjffjcyko tbv Vofxlctt depoh


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


Eoi xicbg Gn sjske esb Uowqbim lsnal iy gvcuti tts itaqwufeb uiq Txdkjynxb axk ryqhfvk Eaggjdmh ecq Jzygvopmi krxuys


Mor tzgwb rwXmkuuajgkj pywvbryl jsp revmzlfyf lmngvetpvmh

Qkgjy ibs


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


Ahveokj


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


Hgpvc hrb


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


Leuvmuw


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


Jbioe gnp


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


Rqcbsxr


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


Jtptr ewc


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


Ots Dkrjacgxm fim ljeuai Avhggdnlz uhcdmlo


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


gbpjd


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


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