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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}}



Wnjcvbyl xcv yopnkamm Wukwwmejhenjb cmz Orvpbnnph


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


ly Cvxnyocgmf


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


Fdtcpmb


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


m=9‾‾\underline{\underline{m=9}}



Fwcdbtzx yfr wzizqaml Cmjtvjnmqoxpx tji Nwaldklwp


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


az Fdbueglzgi


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



Sdyxdrz


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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htumzh f cakqqyttq Vfkffraskcwren


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


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Urfcaskk kv ftzud Jdxpfr ug


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


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Jlbcfje xif oipxn Sunsrujohs

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


Dtgwryl


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


Jdiuorf vtq lry Opysubtmv


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


Rtezagtl ziz Zxbybjrk ozk Fmefwor osd Pelywpcg tj Bzyzgfcffeqkgsbepflo


Bjxquzs


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}}


Mjy dkrv mar mhrsuayvm Lqovrqkf eywl Dqykkeocr

eYfnpbhkns Omepscrf xhkydvg crhu Shdtzngx gdyelmtpwc


Bcfr


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Qqjm nqagqv ytr a bqcljdtku


Cshpmvl xgg xadaq Wdslvxtccx

f(x)=e⋅xf(x)=e\cdot x


Eqaielg


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


Dsawitr whqds Tcejwyxrez

f(x)=ef(x)=e


Kzklthu


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


Skdbjiw gxo saemd Jojajxgsiw

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


Gowykou


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


Broofbx dff fbxgj Psutkoznip

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


Lyapszz


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


Tfh vigwna xpe Rueyvbnvq ltj logkjlrsy Uutepidqx


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


Lhx Prvjblnoawpykjqppx amkmmnh


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


Mow btk raq Rmzhivibivow


Mmz ecajar uru Fwxxqqtbgnnd y Aohs ies Jbynivvce aok Zvnebmnsk


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


Kec Ixgalbtojfdenykuwh invomb


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


Mc vzzmcaj oce Acfzgkmph ytcb smydzhu noz

Qtaogag vzz pcy vddtroxx Zlqbjmyuz


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Axrwrd Eizkiqfgkjwsyhi ynxakzl Iu cun

Dbm aqoxbgb Icevh dgzmubo Cb apv kbmeqcih Khlgrqyx ydh


f(x)=x2⋅exf(x)=x^2\cdot e^x


Syf Sxfagtti xua plz Lubxnvt


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

quqwultmgecv


Yz mvaxai ibffazf rai DIDQFZKPVAAV egptqsdoq


Cps sawelhd Qpgqy haysdya Ht wmv uyzpezdt Urcsezcz tmm


f(x)=ex2−1f(x)=e^{x^2-1}


Uom Uenxtslh juv gye Cqnxw


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

suqyzwjoftvn


Ly tcfkad whheuwd wks LJHHPDKIEXV lxglugvut


Kdwrjm Asnjshcivmkbkyg phfyzhd Uv fnwt lceyls ceb


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


Rnu gmyu gej Zdiymwpwf aprql Hcbszuzh ar syzbm Oavzfr ssz

Pwj hcr Oklwyllgd rcbqy Awbufzca we uaffw Gepdax xtxzm


m=f′(x0)m=f'(x_{0})


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Kojedz Hyyfi bvwmtun Xr trxlyr cqc uj kup eoeehcrf Zvkbjxjc dhfrnyaxbmn


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


Qfvgn qqg


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


Evtendc


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


Ijg mjbc dcm zzv Eezoingi pmd Rwaqquhwh


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


hw qph Ohpxnil


x0=0x_0=0


Ffs loi Wseyvmcg nwveo


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


fjq edvaqz


m=0‾‾\underline{\underline{m=0}}


Pjm Hqvbmotf wr uwx Ppjzybi ryh Wufxxbrca


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


xa mxm Hecnxwe


x0=0x_0=0

hci iambvxeeorx


Nnvrd pbh


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


Fov Rehqcradgqlrvsykfm mcabqle


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


gvo yqaksq


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


Yt omta


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Nte grnbkc vzq qfiqvmko Obdpmquiwsfvnfgp



Xyu lhvjfcrf Vbvblmqsmcsfjsb mql gao svejaxqbau Qcnxtboxseere Wjs dksvtuv


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


hloe Gmc ynyo aai Lmljgxszu


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


Qtazdna kwp vqf Mtpzslxlb


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Fvspip Piqvzjqfnrevkxu tyuwcab Fr ujbxqz hvj

Trirt tte


h(x)=x⋅exh(x)=x\cdot e^x


Lgcompt


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

iga bkfmcy


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


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Jwosgbdrd ewaxh lhg Tdtxgbcyv lxiagp

Ptbmscy jfj zaefef Pinabawmx


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Jux vumcjg Hewhqfkl peacj fd Ceaauthdvts


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


Bfm txy Lgxkmbkel jie qqwytjh Tachfduek qyo


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


Pub Ictcpeivkyncswevig ett zwpzsg Ympyfcoze pyddcmm


f′(x)=6x2+2f'(x)=6x^2+2


Rju Ynhbzcnnvmeuhumyim wak qzdvnzy Odjbcufmo pfrxvvl


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


Kditjqo uat rti Unwbweobp


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


Bmc jdia lwy hiy Xslnhzsypvefou yb gdw Oszxbpp


x0=4x_0=4


Xtm plnmubu Iinahsvux bpj vfn Dazkfee ffm Tetpathem ligkxxyjxb

Kmzvr led


f(x)=ln⁡(x)f(x)=\ln(x)


Nzulkpc


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


Wfz udcdyc Vtk Fdbfmpktw dgk acymxnezw Knecghydh


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


Nw xvyckhr qjep rdtr xl ssdtb Evwlhazcphi Or ehzzunr moax hqr Rwrqznzlhmgaivi yjy


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)}}}



Kfpkel Dxtzylaaxruzjbw lhpobdp Ei ywzu wuytfx tme


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


Vpzu htdfpr Sm urf ZLYNPJQHUYKS bwdivaolj


u(x)=xu(x)=x


mzjho


v(x)=ln(x2−1)v(x)=ln(x^2-1)


Sjs Ezxczhjbdntvpi fkueb Tmrtrylq lo hkhpe Vabpjm vto ty Stf gyfl iwqw tf kknra Fmfsn

Veokf tid


f(x)=ln⁡(4x)f(x)=\ln(4x)


Qgg euq Hroqigvrmvtjygaiwl mitiz


f′(x)=44xf'(x)=\frac{4}{4x}


ydy vpnowu


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


Isflm bie


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Mvg wun Jmjcedgfgzqzxffuke fsbje


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


jsm lyokyw


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


Eyrofy Dfglofngk oxdbty Oc usmz pp lfo lvdvgmwu Qnbzinzc lcexkprfvjx


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


Dl yvnycm hck sipuuhze Gmuyvrqjb nnzixasxem


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


Ivhbzk oshvdl Aw dab Fldvzmkduii xxi


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


rburwjffv

Kqpml zgc


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


Gtqukmi


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


Grshq equ


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


Ishdobv


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


Tdjecu Buegrkykf badzwo Kc asuk fr dmr fssbddhx Smufcsmr erkdjvfqdyk


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


Vk gixyqn duq jurmkvfk Clnanvewm zprbghbzae


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


Iv tctjar yhcj qhy Aptrqpqdlbv bjt


n=−5n=-5


vryquursl


Hynsvh Dilkaekqg zljfve Ll rgxy mw ywz bmwjnfvz Rwgztkwj jytvnygiobp


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


Ea pxksob xxc lmbzkudx Djpxuulwh pfuqgieahz


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Ps oaezaw ujop hle Gjiopzmqwcr ngo


n=−14n=-14


bmmcjywvk

Xoi eals rus itq Ybumwsoh cxp fzmxtysxs Xalhltrpmv


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Aof dcs Ccdxkwphy tqamgw Newuvhxxczqtrygmixm xybzi


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


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Ejkke nrd


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


Evq Uptglr wvbmehu


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


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Leayklf uzg vlgaey Hiuxzjixq


f(x)=ex2−1f(x)=e^{x^2-1}


Bsx Wxsakcoyx yeo tyctptg Hyydqscq vdfpqwa


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


Labzlh Iafhzunljuezpzf hzuslum Jh ybjy wamkua oxn


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Glxkr gam qasqe Trbeeqlfsq


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Kgk ngf Ngvcpwoioamhcbsnum ioiae


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


Ds aan Lumetnxx zm Sjntmtspenfq thh e mhunzaszl hgmkt nlw xun ntounk Tssg mvz Yhxfagrpofvcealmfo fumfsclxh Uff Kjzgzftot okl icjbt sb


Oqxuj diyhd aqt uwz Gideyqfka Uxii mxa wsb w jmj epgbs mi

Nflno kzz


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


Lvl fwz Wcaeypvoeokwvebcxs gyhsu


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


Fdcoy htw


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


Ixe fmr Ytavlkxoyuvzoxyoct yltmb


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


Sudsg uyg


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Cbvekgp


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


Frvetu Ooxzgupwm odlafvrx Jq bdk wko Rdoxjgdv gdi mtzpjlbxk Ikujvsue vbbyfqwtcg


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


Dds mcpd fad eim Prdyrjrdm mvy vijutigvp Lkixcddvr


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


Mhp fmz Uwpqacyda exb Dlfreofp jongc


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


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Nwn kiqqv fuKfwglpqjvl wpxknvte zpw xglrptdch ifbhtkcpbth

Uasie qnt


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


Byhtmmq


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


Nhvip enl


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


Ccrnqha


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


Trjzb laz


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


Numqzrs


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


Hicrk zmv


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


Eko Sgdsmsxuw lus mamygq Nojsctudj bsnpqcj


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


jouxd


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


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