Find the derivative of $y={1over2}tan^4x-{1over2}tan^2x-lncos x$












1















What is the derivative of
$$y={1over2}tan^4x-{1over2}tan^2x-lncos x$$




What I did here was:



$$begin{align}y'&=({1over2}tan^4x-{1over2}tan^2x-lncos x)'\&={1over2}cdot4tan^3x(tan x)'-{1over2}cdot2tan{x}(tan x)'-{1overcos x}(cos{x})'\&={2tan^3xovercos^2x}-{tan xovercos^2x}+{sin xovercos x}\&={2{sin^3xovercos^3x}overcos^2x}-{{sin xovercos x}overcos^2x}+{sin xovercos x}\&={2sin^3xovercos^5x}-{sin xovercos^3x}+{sin xovercos x}\&={2sin^3x-sin xcos^2x+sin xcos^4xovercos^5x}\&={sin x(2-cos^2x+cos^4x)overcos^5x}end{align}$$



I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?



I made a mistake in the beginning while writing down this problem, it should be ${1over4}tan^4x$, I changed it out and got exact results






calculus derivatives trigonometry proof-verification







share|cite|improve this question




















  • 2




    5th line it should be $cos^5x$
    – Yadati Kiran
    Nov 26 at 16:03








  • 2




    last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
    – Yadati Kiran
    Nov 26 at 16:06












  • How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
    – Aleksa
    Nov 26 at 16:10












  • Ah yes I see, I made a mistake in the last step
    – Aleksa
    Nov 26 at 16:12
















1















What is the derivative of
$$y={1over2}tan^4x-{1over2}tan^2x-lncos x$$




What I did here was:



$$begin{align}y'&=({1over2}tan^4x-{1over2}tan^2x-lncos x)'\&={1over2}cdot4tan^3x(tan x)'-{1over2}cdot2tan{x}(tan x)'-{1overcos x}(cos{x})'\&={2tan^3xovercos^2x}-{tan xovercos^2x}+{sin xovercos x}\&={2{sin^3xovercos^3x}overcos^2x}-{{sin xovercos x}overcos^2x}+{sin xovercos x}\&={2sin^3xovercos^5x}-{sin xovercos^3x}+{sin xovercos x}\&={2sin^3x-sin xcos^2x+sin xcos^4xovercos^5x}\&={sin x(2-cos^2x+cos^4x)overcos^5x}end{align}$$



I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?



I made a mistake in the beginning while writing down this problem, it should be ${1over4}tan^4x$, I changed it out and got exact results






calculus derivatives trigonometry proof-verification







share|cite|improve this question




















  • 2




    5th line it should be $cos^5x$
    – Yadati Kiran
    Nov 26 at 16:03








  • 2




    last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
    – Yadati Kiran
    Nov 26 at 16:06












  • How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
    – Aleksa
    Nov 26 at 16:10












  • Ah yes I see, I made a mistake in the last step
    – Aleksa
    Nov 26 at 16:12














1












1








1








What is the derivative of
$$y={1over2}tan^4x-{1over2}tan^2x-lncos x$$




What I did here was:



$$begin{align}y'&=({1over2}tan^4x-{1over2}tan^2x-lncos x)'\&={1over2}cdot4tan^3x(tan x)'-{1over2}cdot2tan{x}(tan x)'-{1overcos x}(cos{x})'\&={2tan^3xovercos^2x}-{tan xovercos^2x}+{sin xovercos x}\&={2{sin^3xovercos^3x}overcos^2x}-{{sin xovercos x}overcos^2x}+{sin xovercos x}\&={2sin^3xovercos^5x}-{sin xovercos^3x}+{sin xovercos x}\&={2sin^3x-sin xcos^2x+sin xcos^4xovercos^5x}\&={sin x(2-cos^2x+cos^4x)overcos^5x}end{align}$$



I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?



I made a mistake in the beginning while writing down this problem, it should be ${1over4}tan^4x$, I changed it out and got exact results






calculus derivatives trigonometry proof-verification







share|cite|improve this question
















What is the derivative of
$$y={1over2}tan^4x-{1over2}tan^2x-lncos x$$




What I did here was:



$$begin{align}y'&=({1over2}tan^4x-{1over2}tan^2x-lncos x)'\&={1over2}cdot4tan^3x(tan x)'-{1over2}cdot2tan{x}(tan x)'-{1overcos x}(cos{x})'\&={2tan^3xovercos^2x}-{tan xovercos^2x}+{sin xovercos x}\&={2{sin^3xovercos^3x}overcos^2x}-{{sin xovercos x}overcos^2x}+{sin xovercos x}\&={2sin^3xovercos^5x}-{sin xovercos^3x}+{sin xovercos x}\&={2sin^3x-sin xcos^2x+sin xcos^4xovercos^5x}\&={sin x(2-cos^2x+cos^4x)overcos^5x}end{align}$$



I'm not sure if this is the end, could this be further simplified in some way? And is it correct to this point at all?



I made a mistake in the beginning while writing down this problem, it should be ${1over4}tan^4x$, I changed it out and got exact results






calculus derivatives trigonometry proof-verification




calculus derivatives trigonometry proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 16:20

























asked Nov 26 at 16:00









Aleksa

33612




33612








  • 2




    5th line it should be $cos^5x$
    – Yadati Kiran
    Nov 26 at 16:03








  • 2




    last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
    – Yadati Kiran
    Nov 26 at 16:06












  • How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
    – Aleksa
    Nov 26 at 16:10












  • Ah yes I see, I made a mistake in the last step
    – Aleksa
    Nov 26 at 16:12














  • 2




    5th line it should be $cos^5x$
    – Yadati Kiran
    Nov 26 at 16:03








  • 2




    last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
    – Yadati Kiran
    Nov 26 at 16:06












  • How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
    – Aleksa
    Nov 26 at 16:10












  • Ah yes I see, I made a mistake in the last step
    – Aleksa
    Nov 26 at 16:12








2




2




5th line it should be $cos^5x$
– Yadati Kiran
Nov 26 at 16:03






5th line it should be $cos^5x$
– Yadati Kiran
Nov 26 at 16:03






2




2




last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
– Yadati Kiran
Nov 26 at 16:06






last line when you factor $sin x$ you have $2sin^2x-cos^2x+cos^4x$.
– Yadati Kiran
Nov 26 at 16:06














How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
– Aleksa
Nov 26 at 16:10






How did you get to that? I got $2sin^2x+cos^2x+cos^4x$
– Aleksa
Nov 26 at 16:10














Ah yes I see, I made a mistake in the last step
– Aleksa
Nov 26 at 16:12




Ah yes I see, I made a mistake in the last step
– Aleksa
Nov 26 at 16:12










2 Answers
2






active

oldest

votes


















2














You made a mistake, the first denominator should be $cos^2 x times cos^3 x = cos^5 x$ and you should be able to do more factoring then



UPDATE



Note after fixing sine power mistake in the last step you have
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$






share|cite|improve this answer























  • Yes I see just changed it now, before I saw your answer
    – Aleksa
    Nov 26 at 16:04












  • Even after fixing it I still don't know what to do
    – Aleksa
    Nov 26 at 16:06










  • @Aleksa see update
    – gt6989b
    Nov 26 at 16:10










  • I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
    – Aleksa
    Nov 26 at 16:14










  • I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
    – Aleksa
    Nov 26 at 16:21



















2














$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$






share|cite|improve this answer























  • Is there something else to do after that point? The solution from the textbook is $tan^5x$
    – Aleksa
    Nov 26 at 16:19










  • I edited my post, there was a typo at the beginning, got the solution after changing everything
    – Aleksa
    Nov 26 at 16:23










  • Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
    – Yadati Kiran
    Nov 26 at 16:23











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














You made a mistake, the first denominator should be $cos^2 x times cos^3 x = cos^5 x$ and you should be able to do more factoring then



UPDATE



Note after fixing sine power mistake in the last step you have
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$






share|cite|improve this answer























  • Yes I see just changed it now, before I saw your answer
    – Aleksa
    Nov 26 at 16:04












  • Even after fixing it I still don't know what to do
    – Aleksa
    Nov 26 at 16:06










  • @Aleksa see update
    – gt6989b
    Nov 26 at 16:10










  • I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
    – Aleksa
    Nov 26 at 16:14










  • I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
    – Aleksa
    Nov 26 at 16:21
















2














You made a mistake, the first denominator should be $cos^2 x times cos^3 x = cos^5 x$ and you should be able to do more factoring then



UPDATE



Note after fixing sine power mistake in the last step you have
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$






share|cite|improve this answer























  • Yes I see just changed it now, before I saw your answer
    – Aleksa
    Nov 26 at 16:04












  • Even after fixing it I still don't know what to do
    – Aleksa
    Nov 26 at 16:06










  • @Aleksa see update
    – gt6989b
    Nov 26 at 16:10










  • I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
    – Aleksa
    Nov 26 at 16:14










  • I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
    – Aleksa
    Nov 26 at 16:21














2












2








2






You made a mistake, the first denominator should be $cos^2 x times cos^3 x = cos^5 x$ and you should be able to do more factoring then



UPDATE



Note after fixing sine power mistake in the last step you have
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$






share|cite|improve this answer














You made a mistake, the first denominator should be $cos^2 x times cos^3 x = cos^5 x$ and you should be able to do more factoring then



UPDATE



Note after fixing sine power mistake in the last step you have
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 16:10

























answered Nov 26 at 16:03









gt6989b

33k22452




33k22452












  • Yes I see just changed it now, before I saw your answer
    – Aleksa
    Nov 26 at 16:04












  • Even after fixing it I still don't know what to do
    – Aleksa
    Nov 26 at 16:06










  • @Aleksa see update
    – gt6989b
    Nov 26 at 16:10










  • I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
    – Aleksa
    Nov 26 at 16:14










  • I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
    – Aleksa
    Nov 26 at 16:21


















  • Yes I see just changed it now, before I saw your answer
    – Aleksa
    Nov 26 at 16:04












  • Even after fixing it I still don't know what to do
    – Aleksa
    Nov 26 at 16:06










  • @Aleksa see update
    – gt6989b
    Nov 26 at 16:10










  • I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
    – Aleksa
    Nov 26 at 16:14










  • I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
    – Aleksa
    Nov 26 at 16:21
















Yes I see just changed it now, before I saw your answer
– Aleksa
Nov 26 at 16:04






Yes I see just changed it now, before I saw your answer
– Aleksa
Nov 26 at 16:04














Even after fixing it I still don't know what to do
– Aleksa
Nov 26 at 16:06




Even after fixing it I still don't know what to do
– Aleksa
Nov 26 at 16:06












@Aleksa see update
– gt6989b
Nov 26 at 16:10




@Aleksa see update
– gt6989b
Nov 26 at 16:10












I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
– Aleksa
Nov 26 at 16:14




I just checked the results of this from the textbook, the solution is $tan^5x$, so there is more to do I think
– Aleksa
Nov 26 at 16:14












I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
– Aleksa
Nov 26 at 16:21




I got the results, the problem was $1over4$ instead of ${1over2}$ made a typo while writing this down and doing it on the computer
– Aleksa
Nov 26 at 16:21











2














$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$






share|cite|improve this answer























  • Is there something else to do after that point? The solution from the textbook is $tan^5x$
    – Aleksa
    Nov 26 at 16:19










  • I edited my post, there was a typo at the beginning, got the solution after changing everything
    – Aleksa
    Nov 26 at 16:23










  • Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
    – Yadati Kiran
    Nov 26 at 16:23
















2














$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$






share|cite|improve this answer























  • Is there something else to do after that point? The solution from the textbook is $tan^5x$
    – Aleksa
    Nov 26 at 16:19










  • I edited my post, there was a typo at the beginning, got the solution after changing everything
    – Aleksa
    Nov 26 at 16:23










  • Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
    – Yadati Kiran
    Nov 26 at 16:23














2












2








2






$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$






share|cite|improve this answer














$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 16:22

























answered Nov 26 at 16:12









Yadati Kiran

1,698519




1,698519












  • Is there something else to do after that point? The solution from the textbook is $tan^5x$
    – Aleksa
    Nov 26 at 16:19










  • I edited my post, there was a typo at the beginning, got the solution after changing everything
    – Aleksa
    Nov 26 at 16:23










  • Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
    – Yadati Kiran
    Nov 26 at 16:23


















  • Is there something else to do after that point? The solution from the textbook is $tan^5x$
    – Aleksa
    Nov 26 at 16:19










  • I edited my post, there was a typo at the beginning, got the solution after changing everything
    – Aleksa
    Nov 26 at 16:23










  • Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
    – Yadati Kiran
    Nov 26 at 16:23
















Is there something else to do after that point? The solution from the textbook is $tan^5x$
– Aleksa
Nov 26 at 16:19




Is there something else to do after that point? The solution from the textbook is $tan^5x$
– Aleksa
Nov 26 at 16:19












I edited my post, there was a typo at the beginning, got the solution after changing everything
– Aleksa
Nov 26 at 16:23




I edited my post, there was a typo at the beginning, got the solution after changing everything
– Aleksa
Nov 26 at 16:23












Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
– Yadati Kiran
Nov 26 at 16:23




Nothing more. You can clear out the denominator to make it look good. Else the result we obtained earlier should be the last.
– Yadati Kiran
Nov 26 at 16:23


















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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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