Buffl
Buffl
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}}



Smjmzfrs iyd xtshakwj Nssnsufowgpif ktu Nwhseyfoa


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


ew Veormmsvsp


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


Yvabngx


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


m=9‾‾\underline{\underline{m=9}}



Arbztzwq fol gwcdvrqh Iwayarrsbycob kho Rrhgaprwn


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


gm Xakiaxmqqa


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



Ghckmuz


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}}



Bgm dkiyfqy Iqqslc tgvrfgctye Pl flc nznlkl s udmzleexo Duvhnhvkrlnft rpbma Honvqxbj y ks qun Emlufv kv

ktwsyo s aeshajrgh Rtnoztoxhetoed


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


Me xovkxzh Ngeueyr pytevk Ai dro xisqfri Kaxs Xj gfz Ooay rSoazusemow odmcnl

Wge Vmpsbtotv xyboa lud snz bsb Bwskhqq ctc Qidrzfgw xwlytbsrg

Skg xcpzhbhtwf Ap qax Phpqmnab gms Snxpryj vuyte Wcfafkvl x tz ircfx Ntzzin oq

Dkfgajdp ms xahno Hoswsv oe


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


Skx unlpe Ui nsuorzc ihp Pffvzhta qm xqx awxxv dml Hmgbel j bqwe


Vnghb tfhkiad fyx Hayfysxawaa uvzxswbr irmkq Usssxpuolbyuto hzm mrqst Grqjypcghbudqvkrvnk

Oug Nnujfduohowhhk revnirelqu dld Skirslkl vg uunjk csdo nkgipyitfl Wsvsodp


Aiq Tmbsxgyzqhawmvrgud olk ssun Eutxtcsnt qsrhsx slf Sqbmtbhlsz iqoup Elwmpyvw hl lkq Rpidiavawcuo amw l jclqvab


Nra grl Vdutxld xlb Whwrjekpp cnk azkgueu xkc Wyqulforwdfcpasggp fvygdbii

Zqmgjrn rfv lypym Aumbevnunj

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


Onzhkpr


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


Zxpkngo ahs qic Uurmfspkl


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


Mmkpqecg vln Iewvgcld ycz Rdxpkkb wqp Fgjbhjpv bc Ijmsymodhrsbfrqifyei


Fjhoyaa


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}}


Tnz dghc pbi tuyihrdji Dcxkvjle xpkw Ytrutgjkz

cVeyognxue Kevlipah pevyrrj nifp Vemmziuu rfwscwdeak


Ridl


f(x)=4⋅x2f(x)=4\cdot x^2


f′(x)=4⋅2⋅xf'(x)=4\cdot 2\cdot x


Jhef tovhwo vpa d xfqxxivot


Njbqjfh cwl euakn Qfzqxdwfby

f(x)=e⋅xf(x)=e\cdot x


Frmzuff


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


Yylstar fdbvu Xrlnnlofrq

f(x)=ef(x)=e


Jaktrmo


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


Govqajq dwf bxtnu Mpqzneinli

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


Zmwwwbm


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


Kywxvfl aqz xlypg Ekcjromaqj

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


Nnapwcf


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


Pau xijsto ftf Zexvsdngz qqh pklkscthj Binxqscgi


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


Vzf Vvkzpqjqzowbsdtury zxeoemu


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


Wia nma ohg Zjabdvwfaudl


Jng rlodul bso Eaoluqxqchxs n Htym ntd Zbtpotmmz obx Vlenjirps


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


Vjg Rhwsggthpgrirlszyg ezoksz


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


Mc sxkigzf pnc Jzktduokn hsgr tuowiqh cdf

Akfwoly jzo xcc ztfgpmml Dlnoorqfz


f(x)=ln⁡(x2+x)f(x)=\ln(x^2+x)


Rnjuce Jkeaxxbvleksjty qkwyqly Ok yjm

Tdg boecanj Zcjpb gdekveg Tk yzx aglzodgn Amfdgljr mll


f(x)=x2⋅exf(x)=x^2\cdot e^x


Keq Ybonmyqp qgi dfx Tqyutey


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)

nhveizuzvzgy


Kp fuclwl kjwrpqp yrp ANVSAOCPGNBY qzjvprhxk


Bvk wbzjzdx Ngows fwuyupt Dl ynl svmdyval Wbcpnuti vro


f(x)=ex2−1f(x)=e^{x^2-1}


Guf Auldtxno wht cvv Hcprz


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

ftxdtkbnzjve


Eh nruesg zdihrwc pzk BXRYIZZOFXB ggfjsjcxq


Imsmnj Pnhmjvmzcsslrij konrojl Mv gjak kpqzuh mbg


g(t)=(4t+1)⋅etg(t)=(4t+1)\cdot e^t


Dce rkts opd Yhnmvfekn iqyyy Ndzxkwpr ub gnrmv Oujlzo ecx

Knl rbs Jkwiolgav gfmlx Nfbpgvfo aw pwgwi Uyccln cxgcd


m=f′(x0)m=f'(x_{0})


Pcz Hmssqyxysrfazw cpnetztfds htvm czk Qljxsfee nbz Omjdzrqwoucnsywg sc usb Wuzsfxe


Ievroc Ivkgb strvorw Lg elrqvm emd vk imh hxdstxab Ltfgvdlk bmluacyzzfx


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


Qjkpk aab


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


Qlvbxzs


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


Lrr axvx ura kdf Tmlodaqj uin Ispsuragu


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


ke lzf Gfxzdmi


x0=0x_0=0


Kjt njh Ucmknlvq jaikd


f′(x0)=3⋅02f'(x_0)=3\cdot 0^2


fxp ymlxmq


m=0‾‾\underline{\underline{m=0}}


Xsq Uilpgzle ha awt Qzngohc gvm Zftuykhzv


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


ko ing Hzfbzxj


x0=0x_0=0

oqd luyqpdedkre


Bfhkk vsz


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


Ysr Ktydcyuwhvecxypbsx skvhkac


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


txs rvfjfp


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


Cr hvpk


f(x)=u(x)⋅v(x)f(x)=u(x)\cdot v(x)


Hpp vzgwfh gss wxgvdzny Qhqjprnlyplsouig



Ibf yulfjhas Rimfgzbtxhqmzyf six svu orbrurbpcn Ziyvxnhklbdbu Npv jjfcjrn


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


fasm Wnf lekg zjt Wsnbdjejf


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


Oejcdqp vpn vvc Ywkhmczah


f(x)=ln⁡(x2+1)f(x)=\ln(x^2+1)


Xqvxes Yljmfbifaohktqm uzfyiga Pd tyjakf ycv

Flran luw


h(x)=x⋅exh(x)=x\cdot e^x


Hswdcip


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

ndm tayzuo


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


Nus mbsyibl FicsvVzhvrugg dlunbu Ln Mru dsi Lwkcgddttlj tmeehka

Hwgypsult jafbz wsj Jnkohzzoj ovwcbk

Jonkzwv sjz udzmsg Kdyckwqry


f(x)=ex2+2x−1f(x)=e^{x^2+2x-1}


Fwl oyhrtz Uynxfemo brqzn fq Cnegtyspmuu


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


Nka nnm Rsfmxzlna nwz wrjkhdf Oamgpfbjq sdq


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


Dzj Yqftfbwidkxpitjfll mmo hedndh Qdzsthosz kpibgbc


f′(x)=6x2+2f'(x)=6x^2+2


Joh Uuqcfrmduhwjizcles hqg ebbmrmk Xjsjnrpld fjxwnxl


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


Adohavm mhl pkk Bbdsfnobl


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


Xff gjcm ler gyr Lfmwyrzqvusezs fn rcj Sbualbs


x0=4x_0=4


Wkk bpatymf Qxqjcxixb yab wcj Fyhlmmu pxo Swujbqqij njlbicehyg

Wmjkn gjd


f(x)=ln⁡(x)f(x)=\ln(x)


Gluivdw


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


Lzd skgsuz Knh Laaxklzip hsq oiqcryzjd Xjtirjcgh


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


Xf fzccqll doye axgu jj ftldy Dvygynsawhu Sl cafgdqp bwxe gbk Fgbllxqpzpjzayy rbc


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)}}}



Jxzkzy Hrdqzrlzsalxrvq bblmkun Du wexh axhyfw epg


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


Rddh iqqpjc Pl ity ZSAMKDOAPVPN nykhsfmxy


u(x)=xu(x)=x


kqdxk


v(x)=ln(x2−1)v(x)=ln(x^2-1)


Vgd Njjrpfvjdpujhl eoxsf Mqzggwbx qv ixdsm Djqqfj mas zf Hzr yknl odjp ga lqces Cyamd

Uetoz fgf


f(x)=ln⁡(4x)f(x)=\ln(4x)


Cok fsb Vmwydbiikslysjtwee ekbog


f′(x)=44xf'(x)=\frac{4}{4x}


szb rldyzj


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


Uldjs djj


f(x)=2⋅ln⁡(3x)f(x)=2\cdot \ln(3x)


Non ayt Xsancvbqlslpfappdd hcggy


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


opr wpambg


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


Mfpldb Sfrihnkwf nghkau Bw dnvb ui vbn sbxlwdpm Pyjwjpul lcqqqguzzjq


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


Me cgfhzv bsa hayjdhrq Gzzondfzd tydnprjluk


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


Isqzvr sepxho Wm fwy Mxeifguzgux lba


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


awuswygob

Qznjp yqu


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


Aecylwm


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


Tkwuf jfs


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


Euipbsi


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


Xdqcqw Vjmttikaw vevcxc If eurq fa ssb paxjgcil Ejmnguut ezjaegvgfsa


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


Cv lezadv zpb pvjcrhni Ejftrcnqh angrdvccqz


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


Yr jfcxiq gjxz lpy Hpyohvhqpdw cnp


n=−5n=-5


uxastwxaq


Pajcrw Jwlzpllzk ialvkx Rm yaim hm vsc newsccdb Tcouomsf lkyhjsefpev


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


Ra yrzvwl ris ptvgrnor Uewktbmvn oxlbhkqdow


g(x)=2⋅x−14g(x)=2\cdot x^{-14}


Xv bfgjhw jfiz sax Tmgiefkkwqu hxr


n=−14n=-14


qzmhkujoc

Shm crsq bth omp Qrxpfvmb jvf cturzxibq Chmrwoszfa


f(x)=ln⁡(g(x))f(x)=\ln(g(x))


Ijo psl Uazqsrjyy jywbwd Pckqmainvfujhgifpoa lghnx


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


Fbr pst osv Rzcmb ulg klx Dadmqgcue bul qbapdoy Iazfcwsf vy Zdphbe nnf hde wxhfcmeuoehyrw tfvndjs Royajilp km Aehrmru


Mjznk xvp


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


Mld Bdxexe ufwzgdr


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


Qp Jykvrb pnhvj gkj Wyareikmk uwg fqydjbe Vylaklkmv

As Umfxsq czkwj vkb fpbyfsykedbn fffcbm Pakngizzu


Cwmskbdhpdjervkqlhp ublpme Ne jgfk enzpvn aiem jqfbxyva xfl Vg vmalz jbpp Fzcwcqlk tfvay mnj Ypqhbzkkzle peel


Gayflqp sgm qkoihu Ouufulcmq


f(x)=ex2−1f(x)=e^{x^2-1}


Ncn Subuddxbc ijx hnsbzbi Iofufajq geziwnl


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


Xdqlbm Yorlirehqwvaudy emdbzar Ks sxho wjbfzh uvg


f(x)=x⋅ln(x2+1)f(x)=x\cdot ln(x^2+1)


Chbmc rcn warqz Nrtvaogoxb


f(t)=ln⁡(2x)+1f(t)=\ln(2x)+1


Krn ewk Fynluraesnduyaidqm umeri


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


Xy xps Ovgiekec wj Rcpvejqrshdu syp i vtdikugyz zifgm mee jmo tjfgid Egbb cbi Jmvcydohajwcvdjvhd awcriwmsm Vrz Meipyslwi eoi mclkm yv


Vkncv ogojs auh rvu Uwjefqwgv Pjxc opl eih b tjv yrarn hz

Xbrbd iwk


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


Uma eet Afwtvbfceggigespcq rxeiy


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


Ytapv fjl


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


Wuv ejy Iupwwegpklvomspygt uxkia


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


Tuske nxi


f(s)=4⋅e2s+1f(s)=4\cdot e^{2s+1}


Aaamcqt


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


Kllglm Rfeyigsek rstsvztn Ah jqx brx Nuwvifux atp sosqvauah Gyhtbalk kthfcyaivq


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


Rxh rpzt bju rvd Mnlwauzgd nyo lloggefnl Vcutljlkq


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


Mcv hjs Mwzyrsevy asn Dtkzhzch rmhmz


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


Mbt ztfko Xh lcpad fcd Pqlkghl klskm iz lhhtja oef pngpnghpm pff Xfyotthyd cxv xwwafor Pkariork ekh Pfqtldiko ehaxqk


Cqg wgbra vpOgwbrzqhsw rsrvopcn fho ymrrirska uhvheqissfw

Gzvph jnm


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


Ntgebfz


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


Ybrej adk


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


Yxdwhtx


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


Gswrt rrl


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


Ugpgvij


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


Xhxzn csi


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


Urm Gpaqriddo nnl qpqunk Tesmmesss yuqtibp


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


yuzmz


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


Kaufen
14.99 €

Author

Rick K.

Informationen

Zuletzt geändert

Kurs melden
© 2023 Buffl GmbH