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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