Find the derivative of $y={1over2}tan^4x-{1over2}tan^2x-lncos x$
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1
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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
asked Nov 26 at 16:00
Aleksa
31912
add a comment |
up vote
1
down vote
favorite
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
asked Nov 26 at 16:00
Aleksa
31912
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
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
asked Nov 26 at 16:00
Aleksa
31912
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
asked Nov 26 at 16:00
Aleksa
31912
asked Nov 26 at 16:00
Aleksa
31912
edited Nov 26 at 16:20
asked Nov 26 at 16:00
Aleksa
31912
asked Nov 26 at 16:00
Aleksa
31912
asked Nov 26 at 16:00
Aleksa
31912
31912
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$
answered Nov 26 at 16:03
gt6989b
32.7k22351
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
add a comment |
up vote
2
down vote
$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$
answered Nov 26 at 16:12
Yadati Kiran
1,259417
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
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$
answered Nov 26 at 16:03
gt6989b
32.7k22351
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
add a comment |
up vote
2
down vote
accepted
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
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$
answered Nov 26 at 16:03
gt6989b
32.7k22351
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
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$
answered Nov 26 at 16:03
gt6989b
32.7k22351
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
$$
cos^4 x - cos^2 x + 2sin ^2 x
= cos^4 x - 3cos^2 x + 2
= left(cos^2 x - 1right)left(cos^2 x - 2right)
$$
answered Nov 26 at 16:03
gt6989b
32.7k22351
edited Nov 26 at 16:10
answered Nov 26 at 16:03
gt6989b
32.7k22351
answered Nov 26 at 16:03
gt6989b
32.7k22351
answered Nov 26 at 16:03
gt6989b
32.7k22351
32.7k22351
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
add a comment |
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
add a comment |
up vote
2
down vote
$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$
answered Nov 26 at 16:12
Yadati Kiran
1,259417
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
add a comment |
up vote
2
down vote
$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$
answered Nov 26 at 16:12
Yadati Kiran
1,259417
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
add a comment |
up vote
2
down vote
up vote
2
down vote
$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$
answered Nov 26 at 16:12
Yadati Kiran
1,259417
$displaystyle{sin x(2sin^2x-cos^2x+cos^4x)overcos^5x}={sin x(2sin^2x-cos^2x(1-cos^2x))overcos^5x}$ $displaystyle={sin^4 x(1+sin^2x)overcos^5x}=tan^4x(sec x+tan xsin x)$
answered Nov 26 at 16:12
Yadati Kiran
1,259417
edited Nov 26 at 16:22
answered Nov 26 at 16:12
Yadati Kiran
1,259417
answered Nov 26 at 16:12
Yadati Kiran
1,259417
answered Nov 26 at 16:12
Yadati Kiran
1,259417
1,259417
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
add a comment |
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
add a comment |
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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